Talk:Wilson's Theorem/Corollary 2

Generalization?

The statement of Wilson's Theorem is:

$\paren {p - 1}! \equiv -1 \pmod p$ for primes $p$

How does one recover Wilson's Theorem from this 'generalization'?

For reference if we plug in $n = p - 1$ we have:

$a_0 = p - 1, \mu = 0$

hence the theorem asserts that:

$\dfrac {\paren {p - 1}!} {p^0} \equiv \paren {-1}^0 \paren {p - 1}! \pmod p$

which is simply

$\paren {p - 1}! \equiv \paren {p - 1}! \pmod p$

It could technically be called a Corollary. --RandomUndergrad (talk) 16:31, 8 July 2020 (UTC)

Makes sense. Haven't a clue where I got this from now, it was from the early days when we hadn't started citing our source works. I'll rename it. --prime mover (talk) 17:45, 8 July 2020 (UTC)

Aha, found it now, it was in Knuth.

In the notation of exercise $12$, we can determine $n! \mod p$ in terms of the $p$-ary representation, for any positive integer $n$, thus generalizing Wilson's theorem. In fact, prove that $n! / p^\mu \equiv \paren {-1}^mu a_0! a_1! \dotsm a_k! \pmod p$.

This was 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.): $\S 1.2.5$: Permutations and Factorials: Exercise $14$ --prime mover (talk) 18:10, 8 July 2020 (UTC)