Wilson's Theorem/Corollary 2/Proof 2

Theorem

Let $p$ be a prime factor of $n!$ with multiplicity $\mu$.

Let $n$ be expressed in a base $p$ representation as:

$\ds n$

$=$

$\ds \sum_{j \mathop = 0}^m a_j p^j$

where $0 \le a_j < p$

$\ds $

$=$

$\ds a_0 + a_1 p + a_2 p^2 + \cdots + a_m p^m$

for some $m > 0$

Then:

$\dfrac {n!} {p^\mu} \equiv \paren {-1}^\mu a_0! a_1! \dotsb a_m! \pmod p$

Consider the numbers of $\set {1, 2, \ldots, n}$ which are not multiples of $p$.

There are $\floor {\dfrac n p}$ complete sets of $p - 1$ such consecutive elements of $\set {1, 2, \ldots, n}$.

Each one of these has a product which is congruent to $-1 \pmod p$ by Wilson's Theorem.

There are also $a_0$ left over which are congruent to $a_0! \pmod p$.

Thus:

the contributions of the divisors which are not multiples of $p$ is $\paren {-1}^{\floor {n / p} } a_0!$

the contributions of the divisors which are multiples of $p$ is the same as the contribution in $\floor {\dfrac n p}!$

Thus the argument can be repeated on $\floor {\dfrac n p}!$ until the formula is complete.

$\blacksquare$