I've been stuck with proving the following equality: $$\prod_{d\mid n} d = n^{\nu(n)/2}$$ Where $\nu(n)$ is the number of divisors of $n$.
Here's how I've approached the problem so far:
The desired equation will be true $$\iff log(\prod_{d\mid n} d) = log(n^{\nu(n)/2})$$ $$\iff \sum_{d\mid n} log(d) = \frac{\nu(n)}{2} \cdot log(n)$$ We consider the PF of $n = p_1^{a_1} \cdot p_2^{a_2} \cdot...\cdot p_k^{a_k}$, and thus $$\frac{\nu(n)}{2}\cdot log(n) = \frac{1}{2}\prod_{p\mid n}(a_i+1)\cdot \sum_{p_i\mid n} a_ilog(p_i)$$ We pair up each $(a_i+1)$ with its respective $a_ilog(p_i)$ term, ultimately getting the following $$=\sum_{p_i\mid n} (a_i^2+a_i)log p_i$$
This feels so close to the final desired result, but I'm struggling to get the final desired result. I'm having trouble specifically understanding what all the possible terms $\sum_{d\mid n} log(d)$ look like.
No need for the final answer - just hints towards it/additional steps that I should be looking to take would be really helpful!
Thanks.
