For example:
$6$ has this property since proper divisors of $6$ are: $2$ and 3.
From this thread: What does the product of all proper divisors equal to?
My attempt was:
If n = p_1^{a_1} \times p_2^{a_2} \times ... \times p_k^{a_k}
Then n = n^{\frac{\tau(n)}{2} - 1}$. Where $\tau(n) = (a_1 + 1) \times (a_2 + 1) \times ... \times (a_k + 1)
So is it good enough to stop here, or we can express n$ in a better formula?