Product of Divisors

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

Let $D \left({n}\right)$ denote the product of the divisors of $n$.

Then:
 * $D \left({n}\right) = n^{\tau \left({n}\right) / 2}$

where $\tau \left({n}\right)$ denotes the $\tau$ function of $n$.

Proof 1
We have by definition that:
 * $D \left({n}\right) = \displaystyle \prod_{d \mathop \backslash n} d$

Also by definition, $\tau \left({n}\right)$ is the number of divisors of $n$.

Suppose $n$ is not a square number.

Let $p \mathrel \backslash n$, where $\backslash$ denotes divisibility.

Then:
 * $\exists q \mathrel \backslash n : p q = n$

Thus the divisors of $n$ come in pairs whose product is $n$.

From Tau Function Odd Iff Argument is Square, $\tau \left({n}\right)$ is even.

Thus $\dfrac {\tau \left({n}\right)} 2$ is an integer.

Thus there are exactly $\dfrac {\tau \left({n}\right)} 2$ pairs of divisors of $n$ whose product is $n$.

Thus the product of the divisors of $n$ is:
 * $\displaystyle \prod_{d \mathop \backslash n} d = n^{\tau \left({n}\right) / 2}$

Now suppose $n$ is square such that $n = r^2$.

Then from Tau Function Odd Iff Argument is Square, $\tau \left({n}\right)$ is odd.

Hence the number of divisors of $n$ not including $r$ is $\tau \left({n}\right) - 1$.

As before, these exist in pairs whose product is $n$.

Thus:

Hence the result.

Proof 2
The result follows from taking square root on both sides.