Divisibility of n-1 Factorial by Composite n

Theorem
Let $n \in \Z$ be composite.

Then:
 * $n \mathrel \backslash \left({n - 1}\right)! \iff n \ne 4$

where:
 * $\backslash$ denotes divisibility
 * $n!$ denotes the factorial of $n$.

Necessary Condition
We have that $3! = 6$ and that $4$ does not divide $6$.

So in order for $n$ to divide $n - 1$ it is necessary that $n \ne 4$.

Sufficient Condition
Let $n \ne 4$ be composite.

Let $n = r s$ where:
 * $r, s \in \Z_{> 1}$
 * $r \ne s$
 * $r, s < n$

This is always possible unless $n = p^2$ for some prime number $p$.

, let $r < s$.

Then by definition of factorial:
 * $\left({n - 1}\right)! = 1 \times 2 \times \ldots \times r \times \ldots \times s \times \ldots \times \left({n - 2}\right) \times \left({n - 1}\right)$

and so:
 * $n = r s \mathrel \backslash \left({n - 1}\right)!$

Let $n = p^2$.

As $n \ne 4$, we have that $p \ne 2$.

Hence $p > 2$.

Hence:
 * $2 p > p^2$

and so:
 * $2 p \mathrel \backslash \left({n - 1}\right)!$

By definition of factorial:
 * $\left({n - 1}\right)! = 1 \times 2 \times \ldots \times p \times \ldots \times 2 p \times \ldots \times \left({n - 2}\right) \times \left({n - 1}\right)$

and so:
 * $n = p^2 \mathrel \backslash \left({n - 1}\right)!$

Hence the result.