TITLE: Showing there exists infinite $n$ such that $n! + 1$ is divisible by atleast two distinct primes
QUESTION [5 upvotes]: This is a homework question. 
I know there exists infinitely many primes. Let $n = p-1$ and so by Wilson's theorem we know there exists atleast one prime $p$ that divides $n! + 1$. I used wolframalpha and checked for a couple of $n = p-1$ values and all show me that there are in fact two distinct primes and one of them is in fact $p$. 
How can I use this to conclude there exists a second prime $q$, $q \neq p$ such that $q$ dividies $n!+1$

REPLY [3 votes]: Equivalently you can show that $(p-1)!+1=p^n$ has no solution for $p \geq 7$ and $n \in \mathbb{N}.$
Hint: $p^{n-1}+...+p+1=(p-2)!$ and $p-1|(p-2)!$ (since $p-1$ is composite), so what can you say about $n$ then? can you derive a contradiction?