Note, October 30, 2026

Wilson’s theorem says (p-1)! \equiv -1 \pmod p for every prime p. The slickest proof I know: in \F_p^\times every element pairs off with its inverse, except the two elements that are their own inverse, 1 and -1 (since x^2 = 1 means p \mid (x-1)(x+1)). So the whole product collapses to

(p-1)! \equiv 1 \cdot (-1) \equiv -1 \pmod p.

The converse holds too: if n > 4 is composite then (n-1)! \equiv 0 \pmod n.