Product of Divisors of some $n$ proof

You want to show that $\prod_{d\mid n}d^2=n^{d(n)}$. Note that $$\prod_{d\mid n}d^2=\prod_{d\mid n}d\frac{n}d=\prod_{d\mid n}n=n^{d(n)}$$

The solution by Pedro Tamaroff is far more compact, and better. We will give the same proof in a more long-winded way.

First we mention a fact about the private lives of the divisors of $n$. It is easy to see that $d$ is a divisor of $n$ if and only if $\frac{n}{d}$ is a divisor of $n$. It turns out that $d$ and $\frac{n}{d}$ are a couple, or to put it in the language of today, they are partners. In your example, $1$ and $12$ are partners, as are $2$ and $6$, as are $3$ and $4$. Note that the product of any $2$ partnered numbers is $n$.

Then if $n$ is not a perfect square, the divisors of $n$ are divided into couples. If $n$ is a perfect square, then $\frac{n}{\sqrt{n}}=\sqrt{n}$, so in that case all the divisors of $n$ are coupled except for $\sqrt{n}$.

Let us look first at the case $n$ not a perfect square. Then there are $\frac{d(n)}{2}$ couples. The product of the elements in any couple is $n$, so the product of all the divisors of $n$ is $n^{d(n)/2}$.

Now suppose that $n$ is a perfect square. Then there are $\frac{d(n)-1}{2}$ couples plus a solitary individual $n^{1/2}$.

The product of the elements in any couple is $n$, so the product of all the coupled elements is $n^{(d(n)-1)/2}$. Multiply by $n^{1/2}$. We get $n^{d(n)/2}$.

I was playing with the Dirichlet convolution that results from the zeta function and its derivative multiplied together and found this roundabout proof, I ended up here because I was just wanting to confirm that it was correct but found that no one offered this solution so here I am giving it.

$$\sum_{d|n} \log d = \log \left( \prod_{d|n} d \right)$$

However we can rewrite that sum the other way as,

$$\sum_{d|n} \log (n/d) = \sum_{d|n} \log n - \sum_{d|n} \log d$$

Like the snake biting its own tail for using integration by parts to integrate $e^x \sin x$ we can add that sum itself to the left hand side and divide by 2 to get

$$\sum_{d|n} \log d = \frac{\tau(n)}{2} \log n$$

Since this is equal, to the original sum, equating them gives the result

$$n^{\tau(n)/2} = \prod_{d|n}d$$