proof of Wilson’s theorem

We first show that, if $p$ is a prime, then $(p-1)!\equiv -1(modp).$ Since $p$ is prime, ${\mathbb{Z}}_p$ is a field and thus, pairing off each element with its inverse in the product $(p-1)!={\prod }_{x=1}^{p-1}x,$ we are left with the elements which are their own inverses (i.e. which satisfy the equation $x^2\equiv 1(modp)$), $1$ and $-1$, only. Consequently, $(p-1)!\equiv -1(modp).$

To prove that the condition is necessary, suppose that $(p-1)!\equiv -1(modp)$ and that $p$ is not a prime. The case $p=1$ is trivial. Since $p$ is composite, it has a divisor $k$ such that , and we have $(p-1)!\equiv -1(modk)$. However, since $k⩽p-1$, it divides $(p-1)!$ and thus $(p-1)!\equiv 0(modk),$ a contradiction.