$1+\frac{1}{2} +\frac{1}{3} +...+\frac{1}{p-1} =\frac{a}{b}$ Let $p\gt 3$, be a prime number and  $1+\frac{1}{2} +\frac{1}{3} +...+\frac{1}{p-1} =\frac{a}{b}$ when $a,b\in \mathbb N$ and $gcd(a,b)=1$. 

prove that $p^2|a$.

I proved that $p|a$, but I cant prove $p^2|a$, and my idea is as follow:
By multiplying $(p-1)!$, we get:
$(p-1)!+\frac{(p-1)!}{2}+...+\frac{(p-1)!}{p-2}+\frac{(p-1)!}{p-1}=\frac{a(p-1)!}{b}$ $\,\,\,\,$ $(1)$
So, the number $\frac{a(p-1)!}{b}$ should be integer. consider the left hand of the sum of $(1)$ in the ring $Z_p$. Since in $Z_p$ the inverses are unique and what we have is indeed a rearrangement of all the congruence classes $mod\,p$. So:
$(p-1)!+\frac{(p-1)!}{2}+...+\frac{(p-1)!}{p-2}+\frac{(p-1)!}{p-1}\equiv 1+2+...+p-1\equiv \frac{p(p-1)}{2}\equiv 0$$\,\,\pmod p$. And from here, $p|\frac{a(p-1)!}{b}$, and that's mean: $p|a$. 
I tried to prove the statement by considering the ring $Z_{p^2}$ but I get nothing.
Thanks in advance… 

This is a consequence/version of Wolstenholme's theorem. A proof is available here.
