Theorem 1 (Generalisation of Wilson's Theorem)   For all integers $1 \leq k \leq p-1,\;p\in\bbP \Leftrightarrow (p-k)!(k-1)!\equiv (-1)^k \pmod{p}$

Proof. If $p$ is a prime, then:
and since $(p-1)! \equiv -1\pmod{p}$ (Wilson's Theorem, simply pair up each number -- except and , the only numbers in which are their own inverses -- with its inverse), the first implication follows.

Now, if $p\div (p-1)!(k-1)! - (-1)^k$ , then $p\in\bbP$ as the opposite would mean that $p=ab$ , for some integers $1<a,b < p$ , and so $p$ would not be relatively prime to $(p-1)!(k-1)!$ as the initial hypothesis implies.