Product of Divisors

Theorem

For any integer \(n \geq 1\), the product of the positive divisors of \(n\) is given by

\[ \prod_{k \mid n} k = n^{\frac{d(n)}{2}}\]

where \(d\) is the divisor counting function.

Proof

We consider the product of square divisors and then, write the permuted product using the fact that a \(\frac{n}{k}\) also varies over all divisors of \(n\) as \(k\) does.

\[\begin{align*} \prod_{k \mid n} k^2 &= \prod_{k \mid n} k \frac{n}{k} \\ &= \prod_{k \mid n} n \\ &= n^{d(n)} \\ \implies \prod_{k \mid n} k &= n^{\frac{d(n)}{2}}. \end{align*}\]