Let $p^m|n$, where $n$ is composite and $\geq 6$. Now if $m \geq 3$ then it is obvious that $mp < p^m \leq n$, and so there are at least $m$ factors of $p$ in $(n-1)!$. Hence $p^m|(n-1)!$. If $m=2$, and $p>2$, then both $p$ and $2p$ are less than $p^2$. Hence $p^2|(n-1)!$. If $p=m=2$, then both $2$ and $4$ are less than $n$, as $n \geq 6$. If $m=1$, then as $n$ is composite $p<n$ and hence $p|(n-1)!$. So any prime power that divides $n$ also divides $(n-1)!$. Hence $n|(n-1)!$.

Or alternatively, If $n$ is composite, unless it is the square of a prime, there exist naturals $a,b$ such that $n=ab$, where $a,b<n, a\not=b$. But $(n-1)!=(n-1)(n-2)....a.....b...2$ so we are done. If $n=p^2$, $(n-1)!=(p^2-1)...p(p-1)...p(p-2)....p$ and so for p>2 clearly is divisible by $p^2$.

Yeah, that's much nicer.

We write the canonical representation of $n$. $n=p_{1}^{a_{1}}p_{2}^{a_{2}}....p_{k}^{a_{k}}.$ Now $n-1>n-p_{1}^{a_{1}}=p_{1}^{a_{1}}. something$ Similarly $n-1>n-p_{k}^{a_{k}}=p_{k}^{a_{k}}.something$ Hence we see that $n|(n-1)!$

r1234 wrote: We write the canonical representation of $n$. $n=p_{1}^{a_{1}}p_{2}^{a_{2}}....p_{k}^{a_{k}}.$ Now $n-1>n-p_{1}^{a_{1}}=p_{1}^{a_{1}}. something$ Similarly $n-1>n-p_{k}^{a_{k}}=p_{k}^{a_{k}}.something$ Hence we see that $n|(n-1)!$ This doesn't work when $n$ is a prime power

A nice extension is found here

Suppose $n=\prod p_i^{e_i}$ where $p_i$s are prime and $i>1$. Then $p_1^{e_1},p_2^{e_2},\dots$ are distinct numbers less that $n-1$. So their product, i.e. $n$, divides $(n-1)!$. If $n=p^k$, $k>2$ and $p$ prime, then $p,p^{k-1}$ are distinct and less than $n-1$, so done. If $n=p^2$ then $p,2p<n-1$, so $n\mid 2p^2\mid (n-1)!$.