Let $n \in \Bbb N_{>1}$. Prove that if $n$ divides $(n − 1)!$, then $n$ is composite. I realize I should use contradiction here, so that I could use Euclid's Lemma and find something, but I'm still struggling, anyone help please?
 A: You have the right idea.
Take a look at what happens if $n$ is prime:
Assume $n$ is prime. If $n$ divides $(n-1)!$, all of its factors must be represented in the factorial somewhere. Since it's prime, it only has 1 factor (itself). Thus, the term $n$ must directly be in the expansion of the factorial. This is a contradiction, since all the terms of the factorial must be less than $n$ by the definition of the factorial. Hence, our assumption was wrong, and $n$ is composite.
A: Hint : By Wilson's theorem , $(p-1)! + 1$ is divisible by $p$ if and only if $p$ is prime.
Hence $(p-1)! $ is never divisible by  a prime $p$.
Can you complete it now ?
For more information :

Wilson's Theorem

A: Assume $p:= n$ is prime.
If $p$ divides $(p-1)!$ then by Euclid's lemma:
$p|(p-1)$ or $p|(p-2)$ or.........$p|(p-(p-2))$ or $p| (p-(p-1))$.
Impossible since $p >(p-1)$, $p >(p-2)$......,$p>1$, a contradiction .
Hence $n$ is composite.
A: The contrapositive is

If $n$ is prime, then $n$ does not divide $(n-1)!$

which is a direct consequence of Wilson's theorem.
