Some convincing reasoning to show that to prove that Ramanujan tau function is multiplicative is very difficult
Before the rest of this answer, I should say that there is essentially no way to show that any problem is hard, except perhaps showing that it implies or reduces to some other problem which people have deemed (for one reason or another) hard. Ramanujan presumably thought this was hard because he tried (and failed) to show that it was multiplicative, and many other people presumably thought this was hard because Ramanujan tried and failed.
For context, the year is 1916. This is when Ramanujan presented his paper "On Certain Arithmetical Functions" to the Cambridge Philosophical Society and stated that he empirically believed that $\tau(\cdot)$ was multiplicative.
The study of multiplicative functions wasn't new. But the functions happened to be interesting arithmetical functions and multiplicativity frequently appears in elementary number theory. Functions weren't studied simply because they were multiplicative (though this would could about later). In fact, from reading Ramanujan's paper (and Hardy's and Mordell's papers around the same time), it appears that the term "multiplicative" hadn't even been introduced yet. [I do not know when that term was introduced].
Thus there is no systematic consideration of showing functions are multiplicative at the time. Simple statements of multiplicativity for elementary functions were considered worth remarking --- for instance, Glaisher proved that many functions were multiplicative in "The Arithmetical Functions $P(m)$, $Q(m)$, $\Omega(m)$ in the Quarterly Journal of Mathematics, and this was considered publication-worthy (though the standard of what was publication worthy was quite different at the time, and Glaisher was serving as the editor of the journal so perhaps his decision was what mattered).
For essentially all previously studied multiplicative arithmetic functions, the multiplicative property was established by essentially combinatorial tools. The elementary functions lend themselves to that very nicely. To that end, many of the tools that Ramanujan used to study $\tau(n)$ were combinatorial (or very nearly combinatorial).
We can phrase questions of multiplicativity in terms of Dirichlet series. These were continuing to rise in prominence at the time. We need some additional context. Dirichlet proved his theorem on primes in arithmetic progressions in 1837, involving Dirichlet $L$-functions. Riemann's memoir appeared in 1860, inspiring greater interaction between complex analysis and functions similar to $\zeta(s)$. Hadamard and de la Vallée-Poussin handled some subtle technicalities in Riemann's analysis and finally proved the prime number theorem in 1896. This is to say that in 1916, Dirichlet series and multiplicative functions are not something widely understood and mastered.
In his paper, Ramanujan included a statement boiling down to proving an Euler product for the associated Dirichlet series $\sum \tau(n) n^{-s}$. But the product is different from simpler ones studied, in that it's a degree 2 product.
The zeta function has a representation $$ \sum_{n \geq 1} \frac{1}{n^s} = \prod_p \frac{1}{1 - \frac{1}{p^{s}}},$$ which is the simplest Euler product. This reflects both unique factorization and the fact that the constant function $1$ is multiplicative. The multiplicativity of Dirichlet characters corresponds to the Euler product $$ \sum_{n \geq 1} \frac{\chi(n)}{n^s} = \prod_p \frac{1}{1 - \frac{\chi(p)}{p^{s}}}.$$ Perhaps the simplest degree 2 Euler product comes from the divisor function $d(n)$, whose Dirichlet series factors as $$ \sum_{n \geq 1} \frac{d(n)}{n^s} = \prod_p \left(\frac{1}{1 - \frac{1}{p^{s}}}\right)^2 = \zeta(s)^2,$$ but this is admittedly pretty simple.
The proposed Euler product (and thus the proposed structure of the multiplicativity) for $\tau(n)$ was $$ \sum_{n \geq 1} \frac{\tau(n)}{n^s} = \prod_p \frac{1}{(1 - \frac{\tau(p)}{p^s} + \frac{p^{11}}{p^{2s}})}.$$ Each individual factor in the Euler product is a degree 2 polynomial in $p^{-s}$. On the level of multiplicativity, this corresponds to the fact that $\tau(\cdot)$ is not completely multiplicative (as opposed to $1$ or $\chi(\cdot)$). And unlike functions like $d(\cdot)$, it doesn't appear that $\tau(\cdot)$ is built out of simpler multiplicative functions. For reference, one classical way to show that $d(\cdot)$ is multiplicative is to show that the convolution of multiplicative functions is multiplicative, and then to show that $d(n) = 1 * 1(n)$ (this is equivalent, of course, to the identity between Dirichlet series $\sum d(n) n^{-s} = \zeta^2(s)$).
Thus the structure of the multiplicativity of $\tau(\cdot)$ is a bit different to previously studied multiplicative functions, and it took a new idea. In short, it is in this context that I would say that proving that $\tau(\cdot)$ is multiplicative is hard.
But it's not actually that hard. Less than a year after Ramanujan's paper appeared, Mordell showed that $\tau(\cdot)$ was indeed multiplicative in his paper "On Mr. Ramanujan's Empirical Expansions of Modular Functions." The idea is new, but simple, and the proof is essentially done in the first 3 pages of the article.
But the idea was new, which is the hardest part. It is interesting to place this in context as well. At the end of his paper, Mordell notes that proving corresponding Euler products for modular functions which are modular on subgroups of $\mathrm{SL}(2, \mathbb{Z})$ are harder, and in particular "it seems hardly worth while to go into details." But later Hecke would go into these details and study what we now call Hecke operators (which essentially standardize Mordell's approach), and show that there are bases of modular forms which are simultaneous eigenfunctions of all the Hecke operators, which in turn implies that they are multiplicative and have a degree $2$ Euler product like $\tau(\cdot)$. And this in turn has been generalized to modular forms on $\mathrm{SL}(3, \mathbb{Z})$, and indeed $\mathrm{SL}(n, \mathbb{Z})$. And so on.
In hindsight, the multiplicativity of $\tau(\cdot)$ is the simplest example of multiplicativity in a large family of modular functions.
Here is some information which indicates that the multiplicative property of the Ramanujan $\tau$-function is not considered as that difficult. When looking at the Wikipedia -article we see the third point stating that for prime $p$ \begin{align*} |\tau(p)|\leq 2p^{\frac{11}{2}} \end{align*} is denoted as Ramanujan's conjecture and this is the part which is difficult.
G.H. Hardy the great mathematician and mentor of Ramanujan wrote in
chapter X: Ramanujan's function $\tau(n)$ of his book Ramanujan - Twelve lectures on subjects suggested by his life and work (1940):
... and $\tau(n)$ is defined by \begin{align*} g(x)=x\{(1-x)(1-x^2)\cdots\}^{24}=\sum_{1}^\infty\tau(n)x^n. \end{align*} I shall devote this lecture to a more intensive study of some of the properties of $\tau(n)$, which are very remarkable and still very imperfectly understood. We may seem to be straying into one of the backwaters of mathematics, but the genesis of $\tau(n)$ as a coefficient in so fundamental a function compels us to treat it with respect.
He continues somewhat later:
- Ramanujan conjectured that \begin{align*} \tau(nn^\prime)=\tau(n)\tau(n^\prime) \end{align*} if $(n,n^\prime)=1$, i.e. that $\tau(n)$ is multiplicative; and this was proved a little later by Mordell. Mordell's proof is very instructive, and sufficiently simple for insertion here. It depends on the identity \begin{align*} \sum_{1}^{\infty}\tau(pn)x^n=\tau(p)\sum_{1}^\infty\tau(n)x^n-p^{11}\sum_{1}^\infty \tau(n)x^{pn}, \end{align*} where $p$ is prime.
Hardy explicitly denotes Mordell's proof, that $\tau(n)$ is multiplicative as both, instructive and sufficiently simple and he presents it in the following two pages.
On the other hand he continues later on in the section The order of $\tau(n)$:
I return to the problem of the order of $\tau(n)$, which I referred to at the end of Lecture IX. This is the most fundamental of the unsolved problems presented by the function.
Ramanujan conjectured that \begin{align*} \tau(p)\leq 2p^{\frac{11}{2}} \end{align*} for every prime $p$. ... I shall call this the Ramanujan hypothesis.
It is this inequality which Hardy considers to be the fundamental part and which he denotes as Ramanujan hypothesis. He continues with:
- We saw in 9.18 that \begin{align*} \tau(n)=O(n^8) \end{align*} Ramanujuan gave a more complicated proof of \begin{align*} \tau(n)=O(n^7) \end{align*} and this is the most that has been proved by elementary methods. I proved in 1918, by the method used by Littlewood and myself in our work on Waring's problem, that \begin{align*} \tau(n)=O(n^6) \end{align*} Kloosterman proved in 1927 that \begin{align*} \tau(n)=O(n^{\frac{47}{8}+\varepsilon}) \end{align*} for every positive $\varepsilon$; Davenport and Salié proved independently in 1933 that \begin{align*} \tau(n)=O(n^{\frac{35}{6}+\varepsilon}) \end{align*} and finally Rankin proved in 1939 that \begin{align*} \tau(n)=O(n^{\frac{29}{5}+\varepsilon}) \end{align*} the best result yet known. The indices here are (apart from the $\varepsilon$'s) less than $6$ by $\frac{1}{8}, \frac{1}{6}$ and $\frac{1}{5}$ respectively.
This was the status of the Ramanujan hypothesis at 1940, the date when the book was published. We now know (see the Wiki-article) that the hypothesis follows from the proof of the Weil conjecture by Deligne (1974).
Conclusion: The iterative improvement of the big $O$ estimate step by step over many years strongly indicates that the Ramanujan Hypothesis is challenging and of another level of difficulty than the multiplicative property of the $\tau$-function.
Addendum [2018-02-25]: The first proof of the multiplicativity of Ramanujan's $\tau$-function by L.J. Mordell was published in the Proceedings of the Cambridge Philosophical Society, vol. XiX dated with 14. June 1917. The contribution was titled On Mr. Ramanujan's Empirical Expansions of Modular Functions.
He introduces the proof by also indicating the level of difficulty (from his point of view of course):
[L.J. Mordell, 1917]: In his paper On Certain Arithmetical Functions Mr Ramanujan has found empirically some very interesting results as to the expansions of functions which are practically modular functions. Thus putting \begin{align*} \left(\frac{\omega_2}{2\pi}\right)^{12}\Delta(\omega_1,\omega_2) =r\left[(1-r)(1-r^2)(1-r^3)\cdots\right]^{24}=\sum_{n=1}^\infty T(n) r^n, \end{align*} he finds that \begin{align*} T(mn)=T(m)T(n)\tag{*} \end{align*} if $m$ and $n$ are prime to each other; and also that \begin{align*} \sum_{n=1}^\infty\frac{T(n)}{n^s}=\prod\frac{1}{1-T(p)p^{-s}+p^{11-2s}}\tag{**} \end{align*} where the product refers to the primes $2,3,5,7,\ldots$. He also gives many other results similar to (**).
My attention was directed to these results by Mr Hardy, and I have found that results of this kind are a simple consequence of the properties of modular functions.
and a few lines later indicating the genius of Ramanujan:
Theorems such as $T(mn)=T(m)T(n)$ had already been investigated by Dr Glaisher for other functions; but the theorems typified by equation (**) seem to be of a new type, and it is very remarkable that they should have been discovered emprically.
The proof of Mr Ramanujan's formulae is as follows. ...
Based on the answer from Markus Scheuer I present the proof for multiplicative property of Ramanujan's Tau function. This is not supposed to be a direct answer to the question here, but it lets the readers judge for themselves the level of difficulty of the multiplicative property of Ramanujan's Tau function.
Let $z\in\mathbb{C}$ be such that imaginary part of $z$ is positive and then the variable $q=e^{2\pi i z} $ satisfies $|q|<1$ and thus we can define a function $f$ by $$f(z) =q\prod_{n=1}^{\infty}(1-q^n)^{24}=\sum_{n=1}^{\infty}\tau(n)q^n\tag{1}$$ The function $f$ is connected to the famous Dedekind Eta function by the relation $f(z) =\eta^{24}(z)$ and consequently satisfies the functional equations $$f(z+1)=f(z),f(-1/z)=z^{12}f(z)\tag{2}$$ Since the transformations $z\to z+1$ and $z\to - 1/z$ generate the full modular group it follows that $$f\left(\frac{az+b} {cz+d} \right) =(cz+d) ^{12}f(z)\tag{3}$$ where $a, b, c, d$ are integers with $ad-bc=1$.
The key is to study another related function $F$ given by $$F(z) =p^{12}f(pz)+\sum_{k=0}^{p-1}f\left(\frac{z+k}{p}\right)\tag{4}$$ where $p$ is a prime number and show that $F$ satisfies exactly the same functional equations given in $(2)$ as satisfied by $f$. The relation $F(z+1)=F(z)$ is easy to verify and we establish that $F(-1/z)=z^{12}F(z)$. Note that expression for $F$ consists of $(p+1)$ terms and replacing $z$ by $-1/z$ just multiplies the first two terms with $z^{12}$. It is bit tricky to show that the same happens for remaining $(p-1)$ terms corresponding to $k=1,2,\dots,p-1$.
For each term in the sum $$\sum_{k=1}^{p-1}f\left(\frac{kz-1}{pz}\right)$$ we can find integers $l, m$ such that $kl+mp=-1$ where $l$ runs through $1,2,\dots,p-1$ modulo $p$ as $k$ runs through $1,2,\dots,p-1$ (this is possible because $k$ is coprime to $p$, see Bezout's identity). Let $z'=(z+l) /p$ and $a=k, b=m, c=p, d=-l$ so that $ad-bc=1$ and $$\frac{az'+b} {cz'+d} =\frac{az'+b} {z} =\frac {az+al+bp} {pz} =\frac{kz-1}{pz}$$ and hence from $(3)$ we get $$f\left(\frac{kz-1}{pz}\right)=f\left (\frac{az'+b} {cz'+d} \right) =(cz'+d) ^{12}f(z')=z^{12}f\left(\frac{z+l}{p}\right)$$ and thus we have proved that $$\sum_{k=1}^{p-1}f\left(\frac{(-1/z)+k}{p}\right)=\sum_{k=1}^{p-1}f\left(\frac{kz-1}{pz}\right)=z^{12}\sum_{l=1}^{p-1}f\left(\frac{z+l}{p}\right)$$ and therefore $F$ satisfies the functional equations as described in $(2)$.
It therefore follows that the function $g$ given by $g(z) =F(z) /f(z) $ satisfies $$g(z+1)=g(z),g(-1/z)=g(z)\tag{5}$$ Next we have \begin{align} F(z) &=p^{12}\sum_{n=1}^{\infty} \tau(n) q^{pn} +\sum_{k=0}^{p-1}\sum_{n=1}^{\infty}\tau(n)q^{n/p}e^{2\pi ikn/p} \notag\\ &=p^{12}\sum_{n=1}^{\infty}\tau(n)q^{pn}+\sum_{n=1}^{\infty}\tau(n)q^{n/p}\sum_{k=0}^{p-1}e^{2\pi ikn/p} \notag\\ &=p^{12}\sum_{n=1}^{\infty}\tau(n)q^{pn}+p\sum_{n=1}^{\infty} \tau(pn) q^n\tag{6} \end{align} and thus $$g(z) =p\tau(p) +c_1q+c_2q^2+\cdots$$ so that $g$ is a modular form of weight $0$ and hence a constant and obviously equal to $p\tau(p) $. We have thus $$F(z) =p\tau(p) f(z) $$ which gives us the identity $$\tau(p) \sum_{n=1}^{\infty}\tau(n)q^n=p^{11}\sum_{n=1}^{\infty}\tau(n)q^{pn}+\sum_{n=1}^{\infty}\tau(pn)q^n\tag{7}$$ The multiplicative property of Ramanujan's tau function is equivalent to the following $$\tau(p^kn) =\tau(p^k) \tau(n) \tag{8}$$ where $p$ is a prime which does not divide $n$. In what follows we show how to get $(8)$ from $(7)$.
The strategy is to apply induction on $k$. For $k=0$ the result is obvious and for $k=1$ the result follows by equating coefficients of $q^n$ in $(7)$. If we equate the coefficients of $q^{p^{k-1}n} $ on both sides of $(7)$ we get $$\tau(p) \tau(p^{k-1}n) =p^{11}\tau(p^{k-2}n)+\tau(p^{k}n)\tag{9}$$ Putting $n=1$ we get $$\tau(p) \tau(p^{k-1})=p^{11}\tau(p^{k-2})+\tau(p^{k})\tag{10}$$ and if we set $$a_k=\tau(p^kn)-\tau(p^k) \tau(n) $$ then the equations $(9), (10)$ lead us to the recursion $$a_{k}=\tau(p)a_{k-1}- p^{11}a_{k-2}$$ Since $a_0=a_1=0$ it now follows by induction that $a_{k} =0$ for all $k$ and the proof is complete.