Solution 1:

I think your proof is fine, but you can take a peek at this paper where it is proved that the numerator of $$H(p-1):=1+\frac{1}{2}+...+\frac{1}{p-1}=\sum_{k=1}^{p-1}\frac{1}{k}$$ is divisible by $\,p^2\,\,,\,p>3$ a prime.

Claims 1-2 in this paper are pretty nice (claim 3 is Geoff's remark), and the proof flows smoothly, imo.