Wilson's theorem and its converse

1. For $p=2,3$ the claim is checked directly ($1!=1\equiv-1\pmod2$, $2!=2\equiv-1\pmod3$). Let $p\ge5$. Each $a\in\{1,\dots,p-1\}$ has a unique inverse $a^{-1}$ in $\{1,\dots,p-1\}$ modulo $p$. We have $a=a^{-1}$ iff $a^2\equiv1$ iff $p\mid(a-1)(a+1)$ iff $a\equiv\pm1$, i.e. $a\in\{1,p-1\}$.

Hence the $p-3$ numbers $2,3,\dots,p-2$ split into pairs $\{a,a^{-1}\}$ with product $\equiv1$, so $2\cdot3\cdots(p-2)\equiv1$, and
$$(p-1)!\equiv1\cdot1\cdot(p-1)\equiv-1\pmod p.$$

2. ($\Rightarrow$) is part 1. ($\Leftarrow$) Suppose $n$ is composite, with a divisor $d$ such that $1<d<n$. Then $d\mid(n-1)!$, so $d\nmid(n-1)!+1$, and therefore $n\nmid(n-1)!+1$. Hence $n\mid(n-1)!+1$ forces $n$ prime.

Remark. Wilson's criterion is a perfect primality test in theory, but useless in practice because $(n-1)!$ is huge.

Lesson. In a group of residues, pair every element with its inverse: only the elements with $a^2=1$ survive.