Product of Proper Divisors

Theorem
Let $n$ be an integer such that $n \ge 1$.

Let $\map P n$ denote the product of the proper divisors of $n$.

Then:
 * $\map P n = n^{\map {\sigma_0} n / 2 - 1}$

where $\map {\sigma_0} n$ denotes the divisor count Function of $n$.

Proof
Let $\map D n$ denote the product of all the divisors of $n$.

From Product of Divisors:
 * $\map D n = n^{\map {\sigma_0} n / 2}$

The proper divisors of $n$ are defined as being the divisors of $n$ excluding $n$ itself.

Thus:
 * $\map P n = \dfrac {\map D n} n = \dfrac {n^{\map {\sigma_0} n / 2} } n = n^{\map {\sigma_0} n / 2 - 1}$