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
Case 1:
If $a=b$ then $n=a^{2}$.
Now $n \mid (n-1)! \implies a \mid (n-1)!$, so
$$\begin{aligned}
(n-1)! &\equiv 1\times 2\times \dotsb \times a \times\dotsb\times (n-a)\times\dotsb\times (n-1) \\ &\equiv 1\times 2\times \dotsb\times a \times\dotsb\times -a\times\dotsb\times -1 \\ &\equiv 0 \pmod n
\end{aligned}$$