Numerator of p-1th Harmonic Number is Divisible by Prime p/Proof 2

Theorem

Let $p$ be an odd prime.

Consider the harmonic number $H_{p - 1}$ expressed in canonical form.

The numerator of $H_{p - 1}$ is divisible by $p$.

Note that for any integer $x$:

$\ds x^p - x$

$\equiv$

$\ds 0$

$\ds \pmod p$

$\ds $

$\equiv$

$\ds x^{\overline p}$

$\ds \pmod p$

$\ds $

$\equiv$

$\ds \sum_k {p \brack k} x^k$

$\ds \pmod p$

By comparing coefficients:

$\ds {p \brack p} \equiv 1 \pmod p$

$\ds {p \brack 1} \equiv -1 \pmod p$

$\ds {p \brack k} \equiv 0 \pmod p$ for $k \ne 1, p$

or in a more compact form:

$\ds {p \brack k} \equiv \delta_{k p} - \delta _{k 1} \pmod p$

where:

$\delta$ is the Kronecker delta.

$\ds H_{p - 1} = \frac {p \brack 2} {\paren {p - 1}!}$

From the above we have:

$\ds p \divides {p \brack 2}$

$p \nmid \paren {p - 1}!$

Hence the numerator of $H_{p - 1}$, when expressed in canonical form, is divisible by $p$.

$\blacksquare$