Wilson's theorem

$p$ prime $\Rightarrow(p-1)!\equiv-1$. Among $1,\ldots,p-1$, pair $k$ with $k^{-1}$. The only self-inverse elements satisfy $k^2\equiv1$, i.e. $k\equiv\pm1$. All other pairs multiply to $1$, so $(p-1)!\equiv1\cdot(-1)\cdot1\cdots1\equiv-1\pmod p$.

Converse. If $n$ composite and $n>4$: $n$ has a proper divisor $d$ with $1<d<n$, so $d\mid(n-1)!$, and $d\mid n$; thus $n\nmid(n-1)!+1$, so $(n-1)!\not\equiv-1\pmod n$. (Small cases: $n=4$, $(3)!=6\equiv2\neq-1$.)

Conclusion. Wilson's criterion holds.

Application. $p=5$: $4!=24\equiv-1\pmod5$. Correct.