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

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

Solution 2:

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.