A simple binomial identity Is there a simple way of showing that a prime $p$ must divide the binomial coefficient $p^n\choose{k}$ for all $n\geq 1$ and $1\leq k\leq p^n-1$?

$$\binom{p^n}{k} = \frac{p^n (p^n -1) ... (p^n -k+1)}{k!}= \frac{p^n}{k} \cdot \frac{ (p^n -1) ... (p^n -k+1) }{(k-1)!} = \frac{p^n}{k} \cdot \underbrace{\binom{p^n -1}{k-1}}_{\in \mathbb{Z}} $$
$p \Big| p^n$ and because $k<p^n$ we get that $$p \Big| \binom{p^n}{k}$$
Q.E.D
