TITLE: If $n\ne 4$ is composite, then $n$ divides $(n-1)!$.
QUESTION [33 upvotes]: I have a proof and need some feedback. It seems really obvious that the statement is true but it is always the obvious ones that are a little trickier to prove. So I would appreciate any feedback. Thank you!
Here is what I am asked to prove:

If $n$ is composite then $(n-1)! \equiv 0 \pmod n$.

Proof:
$n$ is composite $\implies n=ab$ where $a,b \in \mathbb{Z}$ and $0<a,b<n$.
Case 1:
If $a=b$ then $n=a^{2}$.
Now $n \mid (n-1)! \implies a \mid (n-1)!$, so
$$\begin{aligned}
(n-1)! &\equiv 1\times 2\times \dotsb \times a \times\dotsb\times (n-a)\times\dotsb\times (n-1) \\ &\equiv 1\times 2\times \dotsb\times a \times\dotsb\times -a\times\dotsb\times -1 \\ &\equiv 0 \pmod n
\end{aligned}$$
Case 2:
$0<a<b<n$.
Then, since $a \mid n$, $b \mid n$ and $n \mid (n-1)!$ we have that $a \mid (n-1)!$ and $b \mid (n-1)!$.
So this implies
$(n-1)! \equiv 1\times 2\times \dotsb\times a \times\dotsb\times b\times\dotsb\times (n-1) \equiv 0 \pmod n$,
Q.E.D.

REPLY [6 votes]: Recall the definition of the factorial: $$m! = \prod_{k=1}^m k = 1 \times 2 \times 3 \times \dotsb \times m.$$
From this is should be obvious that $ab \mid m!$ for any $1 \le a < b \le m$, since both $a$ and $b$ appear as distinct terms in the product.
In particular, let $n$ be a composite number, and let $m = n-1$.  If $n$ is not the square of a prime, there exist two distinct integers $1 < a < b < n$ such that $n = ab$ (you may want to prove this — it's not difficult), and thus $n \mid (n-1)!$.
What if $n$ is the square of a prime, i.e. $n = p^2$ for some prime $p$?  If $p = 2$, we have a counterexample: $4 \nmid 3! = 6$.
However, if $p > 2$, then $2p < p^2 = n$, and thus we may choose $a = p$ and $b = 2p$ to show that $2n \mid (n-1)!$, and therefore also that $n \mid (n-1)!$.
Finally, the fact that the result does not hold for any prime $n$ follows easily from the fundamental theorem of arithmetic, as the prime factorization of $(n-1)!$ will not contain $n$ if it is prime.  (I'm sure there are weaker lemmas that could be used to prove this, but why bother?  The FToA does it cleanly and easily.)  Thus, for integers $n > 1$, $n \nmid (n-1)!$ if and only if $n$ is either prime or $4$.
(In fact, the result holds trivially for $n = 1$ too, at least under the usual definition of $0! = 1$, but that does not follow from the argument above — $1$ is not composite in the sense needed for the argument.)