linguistic:voice_change, linguistic:translation
Proof using Fermat's little theorem

Again, the result is trivial for p = 2, so suppose p is an odd prime, p ≥ 3. Consider the polynomial

g has degree p − 1, leading term x, and constant term (p − 1)!. Its p − 1 roots are 1, 2,..., p − 1.

Now consider

h also has degree p − 1 and leading term x. Modulo p, Fermat's little theorem says it also has the same p − 1 roots, 1, 2,..., p − 1.

Finally, consider

f has degree at most p − 2 (since the leading terms cancel), and modulo p also has the p − 1 roots 1, 2,..., p − 1. But Lagrange's theorem says it cannot have more than p − 2 roots. Therefore, f must be identically zero (mod p), so its constant term is (p − 1)! + 1 ≡ 0 (mod p). This is Wilson's theorem.