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?
This is impossible due to Bertrand's postulate, since there will always be a prime $ p $ in $ n! $ occuring with multiplicity $ 1 $ as long as $ n \geq 2 $. This actually implies that $ n! $ is never a perfect power for $ n \geq 2 $.
If $\sqrt[n]{n!} = k \in \mathbb{N}$ then $n! = k^n$. When $n\geq 2$ we have $2\mid n!$ so we must also have $2\mid k$ which means that we can write $k = 2^{m} \ell$ for some integers $m$ and $\ell$. This again means that
$$n! = 2^{mn}\ell^n \implies 2^{mn} \mid n!$$
so the power of two that divides $n!$ is $mn$ which is greater or equal to $n$. On the other hand the power of two that divides $n!$ can be computed as
$$\left\lfloor\frac{n}{2}\right\rfloor + \left\lfloor\frac{n}{4}\right\rfloor + \left\lfloor\frac{n}{8}\right\rfloor + \ldots$$
This expression is less than $\frac{n}{2} + \frac{n}{4} + \frac{n}{8} +\ldots = n$ which gives us a contradiction.
If $n\gt1$ then $\sqrt[n]{n!}$ is not an integer (so it is an irrational number). A proof using Bertrand's postulate has been posted. The proof of Bertrand's postulate is somewhat involved. Here is a proof without using Bertrand's postulate.
For a prime number $p,$ the $p$-adic order of a natural number $m,$ denoted by $\nu_p(m),$ is the highest exponent $\nu$ such that $p^\nu$ divides $m;$ the number $m$ is a perfect $k^\text{th}$ power if and only if $\nu_p(m)$ is divisible by $k$ for every prime $p.$ We can show that $n!$ is not a perfect $n^\text{th}$ power (for $n\gt1$) by showing that $\nu_2(n!)$ is not divisible by $n;$ in fact, $0\lt\nu_2(n!)\lt n.$ The lower bound is obvious. For the upper bound, let $m=\lfloor\log_2(n)\rfloor$ and use Legendre's formula: $$\nu_2(n!)=\sum_{k=1}^\infty\left\lfloor\frac n{2^k}\right\rfloor=\sum_{k=1}^m\left\lfloor\frac n{2^k}\right\rfloor\le\sum_{k=1}^m\frac n{2^k}\lt\sum_{k=1}^\infty\frac n{2^k}=n.$$
A much more general (and difficult) result, the Erdős-Selfridge theorem, says that the product of two or more consecutive positive integers is never a perfect $k^\text{th}$ power for any $k\gt1.$
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 $n=1$.
This is (for $n>2$) never an integer, since you always have a single prime number (meaning with exponent 1) and hence its $n$th root, namely the biggest prime smaller than $n$. Since the product of roots of different prime numbers are never integer, this number will never be an integer (for $n>2$).
EDIT: The comment below by bof points out the flaw in my argument. I am not deleting this answer. This proves a much weaker statement version that instead of "never happening" this shows it cannot happen for two consecutive integers.
Bertrand's postulate is the shortest way to prove this as has been suggested in other answers/comments. Here is an alternative attempt.
Suppose $n!=x^n,$ and $(n+1)!=y^{n+1}$ for integers $x,y$. Then we get by dividing these two equations $$ n+1 =\bigg(\frac yx\bigg) ^n y $$ As LHS is an integer so is the RHS. The only way RHS can be an integer is $y/x$ be an integer (and at least 2). That means $n+1$ is bigger than $2^{n+1}$ (because y is at least 2). But this is absurd, giving the required contradiction.
