composite of exponential function
By taking the $i_k$-th term of $\exp \left(\frac{t^k}{k} \right)$ (where $k=1,\ldots, n$), and summing over all such cases where this results in a term containing $t^n,$ we see that the coefficient of $t^n$ in $$ \prod_{i=1}^{\infty} \exp \left( \frac{t^i}{i}\right)$$
is given by $$a_n = \sum_{\ \ \ \ \ \ \ \ \ i_1,\ldots, i_n \geq 0\\ i_1 + 2i_2 + \ldots + n i_n = n} \frac{1}{ 1^{i_1} i_1! \cdot 2^{i_2} i_2! \cdots n^{i_n} i_n!}$$
The lemma below shows that each term in this sum is equal to the probability that a random permutation on $n$ elements is composed of $i_1$ cycles of length $1$, $i_2$ cycles of length $2$, ... and $i_n$ cycles of length $n.$ Therefore, $a_n,$ being the sum of this over all possible cycle length decompositions, is equal to $1.$
Lemma: The number of permutations $\sigma \in S_n$ (the symmetric group on the set $\{1,2,\ldots, n\}$) which have $i_1$ cycles of length $1,$ $i_2$ cycles of length $2,$ ... and $i_n$ cycles of length $n$ is given by
$$ \frac{n!}{ 1^{i_1} i_1! \cdot 2^{i_2} i_2! \cdots n^{i_n} i_n!}$$
Proof: We can construct every such $\sigma$ as follows. Write out the $n!$ permutations of $\{1,2,\ldots, n\}$ and for each permutation, place parenthesis around each of the first $i_1$ terms. Then place parenthesis around pairs of consecutive terms, for the next $i_2$ pairs. Then place parenthesis around triplets of consecutive terms, for the next $i_3$ triplets, and so on. These give all $\sigma$ of the desired cycle length decomposition, however we have overcounted in two ways.
Firstly, disjoint cycles commute (e.g. $(1 \ 3 \ 2) (5 \ 4 \ 6) = (5 \ 4 \ 6)(1 \ 3 \ 2)$), so any of $i_1!$ rearrangements of the $1$-cycles, $i_2!$ rearrangements of the $2$-cycles, ... and $i_n!$ rearrangements of the $n$-cycles results in the same $\sigma.$ Second, for each $k$-cycle we have overcounted by a factor of $k$, by the basic cyclic property of cycles (e.g. $(a \ b \ c) = (c \ a \ b) = (b \ c \ a)$). Dividing $n!$ by the factors of which we overcounted, we get the result.
We can try to do the reverse. Suppose $$\exp(\sum_{n\geq 0}a_nx^n)=\frac{1}{1-x}\tag{1}$$ Putting $x=0$ we can see that $a_0=0$. Differentiating the above equation we get $$\exp(\sum_{n\geq 1}a_nx^n)\sum_{n\geq 1}na_{n} x^{n-1}=\frac{1}{(1-x)^2}\tag{2}$$ and using $(1)$ we get $$\sum_{n\geq 1}na_nx^{n-1}=\frac{1}{1-x}=\sum_{n\geq 1}x^{n-1}$$ so that $a_n=1/n$ for $n\geq 1$.
Since the exponential function is strictly monotone our job is done. Note that all the involved series are having radius of convergence $1$ and hence all the operations are justified for $|x|<1$.
I think this also works in direct manner. Write $$\exp(\sum_{n\geq 1}x^n/n)=\sum_{n\geq 0}a_nx^n$$ and differentiating it we get $$\exp(\sum_{n\geq 1}x^n/n)\sum_{n\geq 1}x^{n-1}=\sum_{n\geq 1}na_nx^{n-1}$$ or $$\sum_{n\geq 0}a_nx^n\sum_{n\geq 0}x^n=\sum_{n\geq 0}(n+1)a_{n+1}x^n$$ and from this one can conclude $a_n=1$.
The approach above uses two keys properties of $\exp(x) $ namely $\exp(0)=1$ and $\exp'(x) =\exp(x) $ and both of these are easily derived from power series definition of $\exp(x) $. Moreover these two properties uniquely characterize $\exp(x) $.