Prove that $\prod_{d \mid n}d=n^{v(n)/2}$ where $v(n)$ is the sum of divisors function.
We have if $n=p_{1}^{a_{1}}p_{2}^{a_{2}} \dots p_{k}^{a_{k}}$ then $v(n)=(a_{1} +1)(a_{2}+1) \dots (a_{k} +1)$ substituting this in the expression does not reach anything, is there any way to express $\prod_{d \mid n}d$ easier to relate to the expression of $v(n)$, or I can proceed to make this demonstration. Thanks
There are two cases to consider, $n$ not a perfect square and $n$ a perfect square.
If $n$ is not a perfect square, the divisors of $n$ can be split into groups of $2$, with product $n$. If $d(n)$ is the number of divisors function, then there are $d(n)/2$ groups of two, and therefore our product is $n^{d(n)/2}$.
A similar argument deals with $n$ a perfect square. We leave it to you to do the details. In this case, the number of divisors function wll be odd, say $d(n)=2k+1$. If $n=m^2$, then one of the terms in our product is $m$, that is, $n^{1/2}$. Use the pairing argument of the preceding paragraph to deal with he other divisors of $n$.
