## pn choose pk and n choose k are congruent mod p

Show that $\binom{pn}{pk} \equiv \binom{n}{k}$ mod $p$.

Note that $(1+x)^{pn} = \sum_{t=0}^{pn} \binom{pn}{t}x^t$ by the binomial theorem. The coefficient of $x^{pk}$ is $\binom{pn}{pk}$. Over $\mathbb{F}_p$, we have $(1+x)^{pn} = ((1+x)^p)^n$ $= (1+x^p)^n$ $= \sum_{t=0}^n \binom{n}{t}(x^p)^t$, and now the coefficient of $x^{pk}$ is $\binom{n}{k}$. In particular, we have $\binom{pn}{pk} \equiv \binom{n}{k}$ mod $p$.