ELI5: Explanation of why the product of divisors of a natural $n$ is always $n^{\frac{d(n)}{2}}$ where $d(n)$ gives the number of divisors I tried reading the answers on Product of Divisors of some $n$ proof and unfortunately couldn't understand them. The accepted answer mentions that
${\displaystyle\prod_\limits{d|n}{d^2}=\prod_\limits{d|n}{d\frac{n}{d}} = \prod_\limits{d|n}{n}=n^{d(n)} }$ where $d(n)$ gives the number of divisors of a number $n$ over $\mathbb N$,
which I assume proves something about the product of (every divisor of $n$ squared) being equal to $n^{d(n)}$. But I still don't get how this is related to the conclusion.
 A: List the factors in increasing order, then again in decreasing order, then multiply the first number in each list, the second etc. There are $d(n)$ calculations that each return $n$, so the overall product is $n^{d(n)}$. The product of $n$'s factors is just the square root of this, since we listed the factors twice.
A: Bearer of the curse, first $d$ is a divisor of $n\iff$ $\frac{n}{d}$ is a divisor of $n$. This is because $d\cdot\frac{n}{d}=n$, and $\frac{n}{d}$ is an integer. This would mean that $$\prod_{d\mid n} d=\prod_{d\mid n} \frac{n}{d}\tag{1}$$ because both products run over the same divisors just in opposite order. Multiplying both sides of $(1)$ by $\prod_{d\mid n} d$ gives $$\prod_{d\mid n} d^2=\prod_{d\mid n} d\prod_{d\mid n} \frac{n}{d}=\prod_{d\mid n} n\tag{2}$$
For the last equality you get an $n$ for each divisor $d$. There are $d(n)$ divisors, we get
$$\prod_{d\mid n} d^2=n^{d(n)}\tag{3}$$
Now, $\prod_{d\mid n} d^2=\left(\prod_{d\mid n} d\right)^2$ so taking the square-root of both sides of $(3)$ yields $$\prod_{d\mid n} d=n^{\frac{d(n)}{2}}$$
