Divisibility of Numerator of Sum of Sequence of Reciprocals

Theorem
Let $p$ be a prime number such that $p > 3$.

Consider the sum of the finite sequence of reciprocals as follows:


 * $S = 1 + \dfrac 1 2 + \dfrac 1 3 + \cdots + \dfrac 1 {p - 1}$

Let $S$ be expressed as a fraction in canonical form, that is:
 * $S = \dfrac a b$

where $a$ and $b$ are coprime.

Then:
 * $p^2 \divides a$

where $\divides$ denotes divisibility.