TITLE: To find all $n$ such that $(n-1)!+1$ is a perfect power of $n$
QUESTION [16 upvotes]: How to find all positive integers $n$ such that $(n-1)!+1$ can be written as $n^k$ for some positive integer $k$?

REPLY [31 votes]: For $n=2,3,5$ the number $(n-1)!+1$ is a perfect power of $n$, but not for $n=1$ or $n=4$. 
Let $n>5$ be such that $(n-1)!+1$ is a perfect power of $n$. If $n$ is composite then $(n-1)!$ is divisible by $n$, so $(n-1)!+1$ cannot be a perfect power of $n$, so $n$ must be prime.
Let $p>5$ be a prime and suppose there exists $k\in\Bbb{Z}$ such that $(p-1)!+1=p^k$. Then
$$(p-1)!=p^k-1=(p-1)\cdot\sum_{i=0}^{k-1}p^i,$$
which shows that $(p-2)!=\sum_{i=0}^{k-1}p^i$. We have
$$(p-2)!\equiv\sum_{i=0}^{k-1}p^i\equiv k\pmod{p-1},$$
and because $p-1>4$ is composite we have $k\equiv(p-2)!\equiv0\pmod{p-1}$.
The inequalities
$$p^k-1=(p-1)!<(p-1)^{p-1}<p^{p-1},$$
of integers show that $k<p-1$, and clearly $k>0$, a contradiction. 
Therefore $n=2,3,5$ are the only positive integers such that $(n-1)!+1=n^k$ for some $k\in\Bbb{Z}$.