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.
 A: 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.)
A: If $n > 4$ and $n = a \cdot b$ with $a, b \geq 2$ then $a + b \leq n - 1$. Since  ${a+b \choose a}$ is an integer it follows that $n = a \cdot b \mid a! \cdot b! \mid (a + b)! \mid (n - 1)!$.
A: Hint $\rm\ n\, =\, \color{#C00}a\:\!\color{#0A0}b\mid1\!\cdot\! 2\cdots\color{#C00} a\:\color{#0A0}{(a\!+\!1)\, (a\!+\!2) \cdots (a\!+\!b)}\cdots (\color{#90f}{ab\!-\!1})=(n\!-\!1)!\,\ $ by $\rm\,\  \color{#0a0}{a\!+\!b}\le \color{#90f}{ab\!-\!1}  $
Note $\rm\,\color{#0A0}b\,$ divides $\rm\color{#0A0}{green}$ product since  a sequence of $\rm\,b\,$ consecutive integers has a multiple of $\rm\,b,\,$
and $\rm\,a\!+\!b \le  ab\!-\!1  \!\!\iff\!\! 2\le (\underbrace{a\!-\!1}_{\large \ge\,1})(\underbrace{\color{#f70}{b\!-\!1}}_{\large\color{#f70}{\ge\, 2}}), \,$ true by $\rm\,a,b\ge 2,\,$ $\rm\underbrace{not\ both\!=\!2,}_{\large n\,=\,ab\,\ne\, 4}\,$ so $\,\color{#f70}{\rm one}$ is $\,\color{#f70}{\ge 3}$
The prior inequality implies that all of the $\rm\color{#0A0}{green}$ factors do occur in $\rm\,(ab\!-\!1)!$
Note $ $ That $\rm\:\color{#0A0}b\:$ divides the above $\rm\color{#0A0}{green}$ term is not deduced from the fact that it is divisible by $\rm\,b!\,$ by integrality of the binomial coefficient $\rm\:(a\!+\!b:a),\:$ as in WimC's answer. Rather, we deduce it from the more elementary fact that a sequence of $\rm\,b\,$ consecutive integers contains a multiple of $\rm\,b,\,$ which is an immediate consequence of (Euclidean) division.
A: 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, $n|(n-1)!$
A: Since $n|(n-1)!$ and $(n-1)!\equiv0\pmod n$ are equivalent, if you can prove one, you should be done.
Take your case 2 for example.  You say "then since $a|n,b|n$, and $n|(n-1)!$..."  But $n|(n-1)!$ is what you set out to prove and you weren't explicit about why it's true, which is the point of a proof in the first place.  
What I think you were going for is that $(n-1)!$ is the product of all positive integers $\le n-1$, which includes $a$ and $b$.  Change the order of multiplication and let the product of all integers $\le n-1$ excluding $a$ and $b$ be equal to a new constant $k$.  So $(n-1)!=kab=kn$.  Therefore, $n|(n-1)!$ or $(n-1)!\equiv0\pmod n$.
Case 1 was a bit more explicit, but you tried using your conclusion to try to prove something else again.  And you missed the loophole that Jonas's comment points out.  You may want to be a bit more explicit about why the product is zero as well rather than just stating that it is.
