= Solution
Write the <composite number> as $n=ab$, with $2\le a\le b<n$. If $a<b$, both $a$ and $b$ occur as distinct factors of the <factorial> $(n-1)!$, so their product $n$ divides that <factorial>.
If $a=b$, then $n=a^2$ and $n>4$ implies $a\ge3$. Both $a$ and $2a$ are distinct factors in $(n-1)!$, because $2a\le a^2-1$. Their product is $2n$, so again $n$ divides the <factorial>. Hence
$$
\boxed{(n-1)!\equiv0\pmod n\quad\text{for every composite }n>4.}
$$
The exception $n=4$ matters: its <factorial> $3!=6$ is not divisible by $4$. This argument handles perfect squares without incorrectly counting the same factor twice.
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