How prove this $(p-1)!\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{p-1}\right)\equiv 0\pmod{p^2}$ 
Show that
  $$(p-1)!\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1}{p-1}\right)\equiv 0\pmod{p^2}.$$

Maybe use this
$$\dfrac{1}{k}+\dfrac{1}{p-k}=\dfrac{p}{k(p-k)}$$
and then I can't. Can you help me to prove it? 
Thank you.
 A: The solution below is adapted from Notes on Wolstenholme’s Theorem by Timothy H. Choi.
Let
$$
S=(p-1)!\sum_{k=1}^{p-1} \frac1k
$$
Using your insight
$$
\dfrac{1}{k}+\dfrac{1}{p-k}=\dfrac{p}{k(p-k)}
$$
we have
$$
2S=(p-1)!\sum_{k=1}^{p-1} \left(\dfrac{1}{k}+\dfrac{1}{p-k}\right) =
p\sum_{k=1}^{p-1} \frac{(p-1)!}{k(p-k)} = pS'
$$
Note that $S'$ is an integer. Now
$$
\frac{(p-1)!}{k(p-k)} \equiv (k^2)^{-1} \bmod p
$$
where the inverse is taken ${}\bmod p$. This is a consequence of Wilson’s Theorem.
Hence
$$
S'\equiv
\sum_{k=1}^{p-1} (k^2)^{-1} \equiv
\sum_{k=1}^{p-1} k^2 = \frac{(p-1)p(2(p-1)+1)}{6} \equiv 0 \bmod p
$$
This means that $2S\equiv 0 \bmod p^2$ and so $S\equiv 0 \bmod p^2$. (We need $p>3$ twice here.)
A: As others have noted, the congruence is not true for $p=3$, since
$$ 2!\left(1+\frac 1 2\right)=2+1=3,$$
which is not divisible by $9$. We can still use what you suggested to prove the congruence holds $\operatorname{mod} p$. Let $p$ be an odd prime, then 
\begin{align*}
(p-1)!\sum_{k=1}^{p-1} \frac 1 k &= (p-1)! \sum_{k=1}^{(p-1)/2} \left(\frac 1 k + \frac 1 {p-k}\right) \\&= (p-1)!\sum_{k=1}^{(p-1)/2} \frac{p}{k(p-k)} = p\sum_{k=1}^{(p-1)/2} \frac{(p-1)!}{k(p-k)}.
\end{align*}
Since $(p-1)!$ always contains $k$ and $(p-k)$ as a factor, the fractions in the sum are integers and the result is a multiple of $p$.
See lhf's answer for why it is even a multiple of $p^2$ as long as $p>3$.
A: I came up with this proof while I work on this problem. 
We write $f(x) \equiv_p g(x)$ if polynomials $f(x), g(x) \in \mathbb{Z}[x]$ satisfies $a_i\equiv b_i$ mod $p$ for all coefficients $a_i$ of $f(x)$, and $b_i$ of $g(x)$. 
Then we have by Fermat's theorem, 
$$
x^{p-1}-1 \equiv_p (x-1)(x-2) \cdots (x-(p-1)).
$$
Group the numbers in pairs as discussed here already, 
$$
\frac 11+\frac 1{p-1}=\frac p{1(p-1)}, \ \ \frac12+ \frac 1{p-2}=\frac p{2(p-2)}, \ \cdots, \ \frac 1{(p-1)/2}+\frac1{(p+1)/2}=\frac p{ (p-1)(p+1)/4 }.
$$
Each group has numerator divisible by $p$. Then it suffices to prove that the numerator $N$ in
$$
\sum_{j=1}^{(p-1)/2} \frac 1{j(p-j)}=\frac N{(p-1)!}\ \ \ (*)
$$
is divisible by $p$.
We may write
\begin{align*}x^{p-1}-1&\equiv_p
(x-1)(x-2)\cdots (x-(p-1))\\ &= \prod_{j=1}^{(p-1)/2} (x-j)(x-(p-j)) \\ &=\prod_{j=1}^{(p-1)/2} (x^2-(j+(p-j))x + j(p-j))\\
&\equiv_p\prod_{j=1}^{(p-1)/2} (x^2+j(p-j))
\end{align*}
Expanding the last product, we find that $N$ is the coefficient of $x^2$. Since $p\geq 5$, the coefficient of $x^2$ in the product must be divisible by $p$. Hence, $p|N$ as desired.
