Editing Wilson's theorem (section)

====Elementary proof====

The result is trivial when <math>p = 2</math>, so assume <math>p</math> is an odd prime, <math>p \geq 3</math>. Since the residue classes modulo <math>p</math> form a field, every non-zero residue <math>a</math> has a unique multiplicative inverse <math>a^{-1}</math>. [[Euclid's lemma]] implies{{efn|Because if <math>a \equiv a^{-1} \pmod{p}</math> then <math>a^2 -1 \equiv 0 \pmod{p}</math>, and if the prime <math>p</math> divides <math>a^2 - 1 = (a - 1)(a + 1)</math>, then by [[Euclid's lemma]] it divides either <math>a - 1</math> or <math>a + 1</math>.}} that the only values of <math>a</math> for which <math>a \equiv a^{-1}\pmod{p}</math> are <math>a \equiv \pm 1 \pmod{p}</math>.  Therefore, with the exception of <math>\pm 1</math>, the factors in the expanded form of <math>(p - 1)!</math> can be arranged in disjoint pairs such that product of each pair is congruent to 1 modulo <math>p</math>. This proves Wilson's theorem.

For example, for <math>p = 11</math>, one has
<math display="block">10! = [(1\cdot10)]\cdot[(2\cdot6)(3\cdot4)(5\cdot9)(7\cdot8)]  \equiv [-1]\cdot[1\cdot1\cdot1\cdot1]  \equiv -1 \pmod{11}.</math>