Product of Divisors

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

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

Then:
 * $\map D n = n^{\map \tau n / 2}$

where $\map \tau n$ denotes the divisor counting function of $n$.

Proof 1
We have by definition that:
 * $\map D n = \displaystyle \prod_{d \mathop \divides n} d$

Also by definition, $\map \tau n$ is the number of divisors of $n$.

Suppose $n$ is not a square number.

Let $p \divides n$, where $\divides$ denotes divisibility.

Then:
 * $\exists q \divides n : p q = n$

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

From Tau Function Odd Iff Argument is Square, $\map \tau n$ is even.

Thus $\dfrac {\map \tau n} 2$ is an integer.

Thus there are exactly $\dfrac {\map \tau n} 2$ pairs of divisors of $n$ whose product is $n$.

Thus the product of the divisors of $n$ is:
 * $\displaystyle \prod_{d \mathop \divides n} d = n^{\map \tau n / 2}$

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

Then from Tau Function Odd Iff Argument is Square, $\map \tau n$ is odd.

Hence the number of divisors of $n$ not including $r$ is $\map \tau n - 1$.

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

Thus:

Hence the result.

Proof 2
The result follows by taking the (positive) square root of both sides.