If $n\ne 4$ is composite, then $n$ divides $(n-1)!$. 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.

We will assume that $n>4$, since $4\hspace{-3pt}\not|\,3!$.
Let $p$ be the smallest factor of $n$. Since $n$ is composite, $p\le\sqrt{n}$.
If $p=\sqrt{n}$, then since $n>4$, we must have $p>2$ so that $2p<p^2=n$. Thus, $p\le n-1$ and $2p\le n-1$, and therefore, $2n=p\cdot2p\,|\,(n-1)!$
If $p<\sqrt{n}$, then $n/p>\sqrt{n}$. Thus, $p\le n-1$ and $n/p\le n-1$, and therefore, $n=p\cdot n/p\,|\,(n-1)!$
In either case, $n|(n-1)!$
