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

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 $n = a \cdot b \mid a! \cdot b! \mid (a + b)! \mid (n - 1)!$.

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.

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.)