'Tis the season!

Nice question (and nice answer too! it provides a very handy trick). Here's a simple proof found out by me.

Product of the divisors of $n$ can be found out as $\large{{ n }^{ \frac { \tau \left( n \right) }{ 2 } }}$ where $\tau \left( n \right)$ refers to the number of divisors of a number $n$ . Here's a nice proof by Calvin Lin on Stack Exchange.

Let the number of divisors be $d$ . So the product of the divisors of $n$ can be written as ${ \large{n }^{ \frac { d }{ 2 } }}$ . We know that product of $d$ integers is $\large{{ n }^{ \frac { d }{ 2 } }}$ . Hence the geometric mean is $d^\text{th}$ root of the number.

The $d^\text{th}$ root is -

$\large{=\sqrt [ d ]{ { n }^{ \frac { d }{ 2 } } } \\ ={ n }^{ \frac { d }{ 2d } }\\ ={ n }^{ \frac { 1 }{ 2 } }\\ =\sqrt { n } }$

Thanks bro!

Yes! Check my comment.