TITLE: When is $(p - 1)! + 1$ a power of $p$?
QUESTION [21 upvotes]: A friend asked me this question:

If $p$ is a prime, prove that $(p - 1)! + 1$ is a power of $p$ if and only if $p = 2, 3$ or $5$.

Clearly one direction is obvious, namely that $p=2,3,5$ implies $(p - 1)! + 1$ is a power of $p$.
The other direction is not clear to me. Since by Wilson's theorem $p$ divides $(p - 1)! + 1$ so we need to show that if there are no other prime factors then $p=2,3,5$. Can someone give me a hint for establishing this?
Thanks

REPLY [5 votes]: Not sure if only one hint would help, you could try using more hints as needed.  First, the statement holds for all natural numbers $n \notin \{2, 3, 5\} $, not just for primes.  One way to proceed is to show that for $n > 5$:
1) $n$ cannot be even if it satisfies the equation
2) for $n$ odd, $(n-1)^2 \space | \space (n-1)!$
3) from (2) above and the given equation, show $(n-1) \space | \space k$ which implies $n-1 \le k$
4) show that $\forall n > 2,  (n-1)! < n^{n-1} - 1$  
Hope that helps!