product of divisors function

The product of all positive divisors of a nonzero integer $n$ is equal $\sqrt{n^{\tau (n)}}$, where tau function $\tau (n)$ expresses the number of the positive divisors of $n$.

Proof.  Let  $t=\tau (n)$  and the positive divisors of $n$ be 

If $n$ is not a square of an integer, $t$ is even (see http://planetmath.org/node/11781parity of $\tau$ function), whence

$\{\begin{array}{cc}{a}_1{a}_t=n\hfill & \\ a_2{a}_{t-1}=n\hfill & \\ \mathrm{\cdots }\hfill & \\ a_{\frac{t}{2}}{a}_{\frac{t+2}{2}}=n.\hfill & \end{array}$

Thus

$$\prod _{d\mid n}d=a_1{a}_2\mathrm{\cdots }{a}_t=n^{\frac{t}{2}}.$$

If $n$ is a square of an integer, $t$ is odd, and we have

$\{\begin{array}{cc}{a}_1{a}_t=n\hfill & \\ a_2{a}_{t-1}=n\hfill & \\ \mathrm{\cdots }\hfill & \\ a_{\frac{t-1}{2}}{a}_{\frac{t+3}{2}}=n\hfill & \\ a_{\frac{t+1}{2}}=n^{\frac{1}{2}}.\hfill & \end{array}$

In this case we obtain a result:

$$\prod _{d\mid n}d=a_1{a}_2\mathrm{\cdots }{a}_t=n^{\frac{t-1}{2}+\frac{1}{2}}=n^{\frac{t}{2}}$$

Note.  The absolute value of the product of all divisors is $n^{\tau (n)}.$