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

Proof
Adding 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 $n = \left({p - 1}\right)!$

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

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

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

We have that:
 * $\dfrac {\left({p - 1}\right)!} k \equiv k' \pmod p$

where $k'$ is determined by:
 * $k k' \bmod p = 1$

The set:
 * $\left\{ {1', 2', \ldots, \left({p - 1}\right)'}\right\}$

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

in a different order.

Thus $m$ is congruent to $\dfrac {\left({p - 1}\right)!} {1 + 2 + \cdots + p - 1} \equiv 0 \pmod p$.