Product of Divisors of some $n$ proof The function $d(n)$ gives the number of positive divisors of $n$, including n itself.
So for example, $d(25) = 3$, because $25$ has three divisors: $1$, $5$, and $25$.
So how do I prove that the product of all of the positive divisors of $n$ (including $n$ itself) is $n^{\frac{d(n)}{2}}$.
For example, the divisors of $12$ are $1$, $2$, $3$, $4$, $6$, and $12$. $d(12)$ is $6$, and
$1 · 2 · 3 · 4 · 6 · 12 = 1728 = 12^3 = 12^{\frac{6}{2}} = 12^{\frac{d(n)}{2}}$

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 $n^{d(n)/2}$. 
