Numerator of p-1th Harmonic Number is Divisible by Prime p/Proof 1

Proof
Add the terms of $H_{p - 1}$ using the definition of rational addition to obtain $\dfrac m n$.

Do not cancel common prime factors from $m$ and $n$.

It is seen that $n = \paren {p - 1}!$

Hence $p$ is not a divisor of $n$.

The numerator $m$ is seen to be:
 * $m = \dfrac {\paren {p - 1}!} 1 + \dfrac {\paren {p - 1}!} 2 + \cdots + \dfrac {\paren {p - 1}!} {p - 1}$

Thus it is sufficient to show that $m$ is a multiple of $p$.

Each term in this sum is an integer of the form $\dfrac {\paren {p - 1}!} k$.

For each $k \in \set {1, 2, \ldots, p - 1}$, define $k'= - \dfrac {\paren {p - 1}!} k \bmod p$.

By Wilson's Theorem
 * $k k' \equiv -\paren {p - 1}! \equiv 1 \pmod p$

Therefore
 * $k' \equiv k^{-1} \pmod p$

From the corollary to Reduced Residue System under Multiplication forms Abelian Group:
 * $\struct {\Z'_p, \times}$ is an abelian group.

Since Inverse in Group is Unique, the set:
 * $\set {1', 2', \ldots, \paren {p - 1}'}$

is merely the set:
 * $\set {1, 2, \ldots, p - 1}$

in a different order.

Thus