TITLE: is $\sqrt[n]{n!}$ ever an integer?
QUESTION [9 upvotes]: Does there exist an $n \in \mathbb{N}$ greater than $1$ such that $\sqrt[n]{n!}$ is an integer?

The expression seems to be increasing, so I was wondering if it is ever an integer. How could we prove that or what is the smallest value where it is an integer?

REPLY [2 votes]: In this answer, it is shown that the number of factors of $p$ that divide $n!$ is
$$
\frac{n-\sigma_p(n)}{p-1}\tag{1}
$$
where $\sigma_p(n)$ is the sum of the base-$p$ digits of $n$.
For $n!$ be an $n^{\text{th}}$ power, $(1)$ must be a multiple of $n$ for any prime $p$.
For any $n\ge1$, we have $\sigma_p(n)\ge1$ for any prime $p$. Thus, $(1)$ is less than $n$, and since it must be a multiple of $n$, it must be $0$.
Thus, either $n=0$ or the number of factors of any prime $p$ that divides $n!$ must be $0$. Therefore, we either have $n=0$ or $n=1$.