Given integers $a \ge b > 0$ and a prime number $p$, prove that ${pa \choose pb} \equiv {a \choose b} \mod p$. I've been grappling with this problem for a while but haven't solved it.
Given integers $a \ge b > 0$ and a prime number $p$, prove that ${pa \choose pb} \equiv {a \choose b} \mod p$.
 A: $$(1+x)^{pa}= \sum_{n=0}^{pa} {pa \choose n} x^{n}$$
\begin{align}
(1+x)^{pa}=\left ((1+x)^{p} \right )^a=\left (\sum_{k=0}^{p}{p \choose k}x^{k} \right )^{a} \Rightarrow\\
\end{align}
\begin{align}
(1+x)^{pa}&\equiv\left (\sum_{k=0}^{p}{p \choose k}x^{k} \right )^{a} \mod p\\
&\equiv(1+x^p)^{a} \mod p\\
&\equiv \sum_{i=0}^{a} {a \choose i}x^{pi} \mod p
\end{align}
Equating coefficients $\mod p$ it yields $$  {  pa \choose pb} \equiv {a\choose b} \mod p$$
A: The following is a combinatorial argument.
Draw $a$ concentric circles and divide them radially into $p$ parts each.  There are then a total of $pa$ regions.  There are $\binom{pa}{pb}$ ways to select $pb$ of these regions.  Consider the action of rotation by $2\pi/p$ on these selections.  There are $\binom{a}{b}$ selections which are fixed by the rotation: these are the selections that consist of $b$ complete annuli.  All others fall into orbits of size $p$.  The desired conclusion $$\binom{pa}{pb} \equiv \binom{a}{b}\, (\text{mod}\,p)$$ 
follows.
