Wilson's Theorem/Corollary 2/Proof 1

Proof
Proof by induction:

Let $\map P n$ be the proposition:
 * $\dfrac {n!} {p^\mu} \equiv \paren {-1}^\mu a_0! a_1! \dotsm a_k! \pmod p$

where $p, a_0, \dots, a_k, \mu$ are as defined above.

Basis for the Induction
For $n = 1$:
 * $a_0 = 1, \mu = 0$

$\map P 1$ reduces to:
 * $\dfrac {1!} {p^0} \equiv \paren {-1}^0 1! \pmod p$

which holds trivially.

This is the basis for the induction.

Induction Hypothesis
This is our induction hypothesis:
 * $\dfrac {r!} {p^\mu} \equiv \paren {-1}^\mu a_0! a_1! \dots a_k! \pmod p$

Now we need to show true for $n = r + 1$:
 * $\dfrac {\paren {r + 1}!} {p^{\mu'}} \equiv \paren {-1}^{\mu'} b_0! b_1! \dots b_k! \pmod p$

where $\ds r + 1 = \sum_{j \mathop = 0}^k b_j p^j$ is the base $p$ presentation of $r + 1$, and:
 * $p^{\mu'} \divides \paren {r + 1}!$
 * $p^{\mu' + 1} \nmid \paren {r + 1}!$

Induction Step
This is our induction step:

Let $p^m$ be the largest power of $p$ which divides $r + 1$, that is:
 * $p^m \divides r + 1$
 * $p^{m + 1} \nmid r + 1$

Then we must have:
 * $b_m \ne 0$
 * $b_j = 0$ for $0 \le j < m$
 * $\mu' = \mu + m$

and consequently:
 * $a_j = b_j$ for $j > m$
 * $a_m + 1 = b_m$
 * $a_j = p - 1$ for $0 \le j < m$

Thus:

By the Principle of Mathematical Induction, $\map P n$ is true for all $n \in \Z_{>0}$.