According to the theorem, if
$$1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\cdots+\frac{1}{p-1} =\frac{r}{q}$$
then we have to prove that $r\equiv0 \pmod{p^2}$.
(Given $p>3$, otherwise $1+\dfrac{1}{2}=\dfrac{3}{2}$, $3 \not\equiv 0 \pmod 9$.)
I guess there's a $(\bmod p)$ solution for this, but I don't really get how to start it.
Best Answer
The group theoretic proof on Wikipedia is good.
Alternatively, another nice way is, the second line is working in $\mathbb{Z}/p\mathbb{Z}$,
$$ \frac2p\sum_{n=1}^{p-1} \frac{1}{n} = \frac2p \sum_{n=1}^{(p-1)/2} \left(\frac{1}{n} + \frac{1}{p-n}\right) = 2 \sum_{n=1}^{(p-1)/2} \frac{1}{n(p - n)}$$ $$ \equiv -\sum_{n=1}^{p-1} \frac{1}{n^2} \equiv - \sum_{n=1}^{p-1} n^2 \equiv -\frac{(p-1)p(2p - 1)}{6}$$
which appears in this article by Christian Aebi and Grant Cairns.
A more hands-on approach can be found in these notes by Timothy Choi.
Finally, an excellent survey and additional results here by Romeo Mestrovic.