Write $1/1 + 1/2 + ...1/ (p-1)=a/b$ with $(a,b)=1$. Show that $p^2 \mid a$ if $p\geq 5$ is prime [Wolstenholme's theorem]

Write $\frac 11 + \frac12 + ...\frac1{(p-1)} =$ $\frac ab$.Such that $(a,b)=1$. Show that $p^2 \mid a$ $\text{if}$ $p\geq 5$.

I was trying to apply something with prime modulus, but I am unsure how exactly to go about it.

Let $f(x)=(x-1)(x-2)\dots(x-(p-1))$. Now, by Fermat, we know that the coefficients of $f(x)$ other than the $x^{p-1}$ and $x^0$ are divisible by $p$.

So if $f(x)=x^{p-1} + \sum_{i=0}^{p-2} a_i x^{i}$ and $p\geq 5$ then $p\mid a_2$, so $$f(p)\equiv a_1p + a_0\pmod {p^3}$$

But we see that $f(x)=(-1)^{p-1}f(p-x)$ for any $x$, so if $p$ is odd, $f(p)=f(0)=a_0$, so that means that:

$$0=f(p)-a_0 \equiv a_1p\pmod {p^3}$$

So $0\equiv a_1\pmod{p^2}$.

Now your sum is just $\frac{a_1}{(p-1)!}$.

Write this as $\sum_{j=1}^{\frac{p-1}{2}} \frac{p}{j(p-j)}.$ Then it suffices to show that $\sum_{j=1}^{\frac{p-1}{2}} \frac{1}{j^{2}}$ is a rational number with numerator divisible by $p.$ Working (mod $p$), twice this last quantity is $\sum_{j=1}^{p-1} \frac{1}{j^{2}},$ which (mod $p$) is the same as $\sum_{j=1}^{p-1} j^{2} = \frac{(p-1)p(2p-1)}{6},$ and this is $0$ mod $p$ for $p \geq 5.$