Binomial Congruence

Question

How can we show that $\dbinom{pm}{pn}\equiv\dbinom{m}{n}\pmod {p^3}$ where $m$ and $n$ are nonnegative integers and $p$ is a prime such that $p \geq 5$ ? I can do it for $\mod p^2$ but I am stuck here.

Answer 1

Since: $$\binom{pm}{pn}=\frac{(pm)(pm-1)\cdot\ldots\cdot(p(m-n)+1)}{(pn)!}=\binom{m}{n}\prod_{k=0}^{n-1}\prod_{h=1}^{p-1}\frac{(m-n+k)p+h}{kp+h}$$ it is enough to show that $$\binom{kp-1}{p-1}\equiv{1}\pmod{p^3}\tag{1}$$ for any $k$ and for any $p\geq 5$, that is also known as Glaisher's theorem. It happens that: $$\binom{kp-1}{p-1}=\frac{((k-1)p+1)\cdot\ldots\cdot((k-1)p+p-1)}{(p-1)!}=\prod_{h=1}^{p-1}\left(1+\frac{(k-1)p}{h}\right),\tag{2}$$ so: $$\binom{kp-1}{p-1}\equiv 1+p(k-1) H_{p-1} + (k-1)^2 p^2\cdot\!\!\!\!\!\!\sum_{1\leq h_1<h_2\leq p-1}\frac{1}{h_1 h_2}\pmod{p^3},\tag{3}$$ but since $H_{p-1}\equiv 0\pmod{p^2}$ by Wolstenholme's theorem, we just need to prove that $$\sum_{1\leq h_1<h_2\leq p-1}\frac{1}{h_1 h_2}=\frac{1}{2}\left(H_{p-1}^2-H_{p-1}^{(2)}\right)\equiv 0\pmod{p},\tag{4}$$ that is trivial since $H_{p-1}^{(2)}\equiv 0\pmod{p}$.