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.
 A: 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 $p \geq 5.$
A: Lets call $s:=(p-1)!(1 +\frac{1}{2} +\cdots +\frac{1}{p-1})$
Since, $s$ contains a even number of terms.
For $i =1,2, \dots, \frac{p-1}{2}$ and let's add them as follows:
\begin{equation*}
\frac{1}{i} + \frac{1}{p-i}= \frac{p}{i(p-i)} \quad \text{for} \quad i =1,2, \dots , \frac{(p-1)}{2}
\end{equation*}
\begin{equation*}
s = (p-1)! \sum_{i=1}^{\frac{p-1}{2}}\frac{p}{i(p-i)}
\end{equation*}
Cleary $p$ divides $s$.
Let $t$ be defined by:
\begin{equation*}
t=(p-1)!\left(\displaystyle \sum_{i=1}^{\frac{p-1}{2}}\frac{1}{i(p-i)} \right)
\end{equation*}
To prove that $t\equiv 0 \quad (\text{mod $p$})$.
Passing to $\text{mod $p$}$, and using Wilson's Theorem one obtains:
\begin{equation*}
t \equiv -\displaystyle \sum_{i=1}^{\frac{p-1}{2}}\frac{1}{i(p-i)}=\sum_{i=1}^{\frac{p-1}{2}}\frac{1}{i^2} =\frac{1}{2} \sum_{i=1}^{p-1}i^2=\frac{(p-1)p(2p-1)}{6}
\end{equation*}
which is clearly divisible by 2.
A: My solution below is not as elegant as that of @ThomasAndrews and is similar to that of @GeoffRobinson. I hope it adds to the understanding of the reader.
The problem is equivalent to showing the following

For a prime $p\geq 5$, the sum $1^{-1}+ 2^{-1}+ \cdots + (p-2)^{-1}+ (p-1)^{-1}$ is $0$ in $\mathbf Z/p^2\mathbf Z$.

We prove this as follows: For each $1\leq i\leq p-1$, write $a_i$ to denote $i^{-1}$. Then note that for each $1\leq i\leq p-1$, $(p+i)^{-1}$ is $a_i+pk_i$ for some integer $k_i$. We show that $\sum_{i=1}^{p-1} k_i$ is divisible by $p$.
To see that, note that we have $(p+i)(a_i+pk_i)\equiv 1\pmod{p^2}$. Using $ia_i\equiv 1\pmod{p^2}$, we get $p(a_i+ik_i)\equiv 0\pmod{p^2}$. This is equivalent to $a_i+ik_i\equiv 0\pmod{p}$. Multiplying both sides by $a_i$, and using $ia_i\equiv 1\pmod{p}$, we have $a_i^2+k_i\equiv 0\pmod{p}$. Thus $$\sum_{i=1}^{p-1}k_i\equiv -\sum_{i=1}^{p-1}a_i^2 \equiv -\sum_{i=1}^{p-1}i^2\pmod{p}$$
But $\sum_{i=1}^{p-1}i^2$ is divisible by $p$ if $p\geq 5$, so the last congruence is $0$.
What we have shown is that 

$\sum_{i=1}^{p-1} i^{-1} = \sum_{i=1}^{p-1} (p+i)^{-1}$ in $\mathbf Z/p^2\mathbf Z$.

More generally one can show that

$\sum_{i=1}^{p-1} i^{-1} = \sum_{i=1}^{p-1} (pt+i)^{-1}$ in $\mathbf Z/p^2\mathbf Z$, for any integer $t$.

So if we write $s$ for $\sum_{i=1}^{p-1}i^{-1}$, we have, by putting $t=p-1$ in the above, that
$$s = \sum_{i=1}^{p-1}(p(p-1)+i)^{-1} = \sum_{i=1}^{p-1}(-i)^{-1}=-\sum_{i=1}^{p-1}i^{-1}=-s$$
Showing $s=-s$ in $\mathbf Z/p^2\mathbf Z$, and thus $s=0$ in $\mathbf Z/p^2\mathbf Z$.
A: 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 $\frac{a_1}{(p-1)!}$.
