TITLE: Show that $(\binom{p^2}{p} -p ) $ is divisible by $p^5$, for every prime number $p, p\ge 5$
QUESTION [7 upvotes]: Show that $(\binom{p^2}{p} -p ) $ is divisible by $p^5$, for every prime number $p, p\ge 5$. 
I have a combinatorics problem, and this is what it reduces to. I am not quite sure how to link the fifth power in divisiblity. 
Edit: I have shown this is equivalent to $$(p-1)!\cdot(\sum_{i=1}^{p-1} \frac{1}{i}) \equiv 0 \pmod  {p^2}$$

REPLY [6 votes]: It's sufficient to prove that $\sum\limits_{i=1}^{p-1} \frac{1}{i} \equiv 0 \pmod {p^2}$.
Denote $S=\sum\limits_{i=1}^{p-1} \frac{1}{i}$. Then, note that
$$
2S=\sum\limits_{i=1}^{p-1} \left(\frac{1}{i}+\frac{1}{p-i}\right)=
p\cdot \sum\limits_{i=1}^{p-1} \frac{1}{i(p-i)}.
$$
Hence, it's sufficient (since $p>3$) to prove that 
$$
\sum\limits_{i=1}^{p-1}\frac{1}{i(p-i)}\equiv 0\pmod p.
$$
However,
$$
\sum\limits_{i=1}^{p-1}\frac{1}{i(p-i)}\equiv -\sum\limits_{i=1}^{p-1}\frac{1}{i^2}\pmod p.
$$
Since modulo $p$ we have $\{1,2,\ldots, p-1\}=\left\{\frac{1}{1},\frac{1}{2},\ldots, \frac{1}{p-1}\right\}$ we obtain 
$$
\sum\limits_{i=1}^{p-1}\frac{1}{i^2}\equiv \sum\limits_{i=1}^{p-1} i^2=\frac{p(p-1)(2p-1)}{6}\equiv 0\pmod p
$$
for prime $p\geq 5$.
Thus, $\sum\limits_{i=1}^{p-1} \frac{1}{i^2} \equiv 0 \pmod {p}$ and so $\sum\limits_{i=1}^{p-1} \frac{1}{i} \equiv 0 \pmod {p^2}$, as desired.