We first show that, if $p$ is a prime, then $(p-1)! \equiv -1 \pmod p.$ 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 $x^2 \equiv 1 \pmod p$ ), $1$ and $-1$ , only. Consequently, $(p-1)! \equiv -1 \pmod p.$

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