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 $1\leq k\leq p^n-1$?

Just a quick remark after the fact: If you accept that  $$ (a +b )^{p} \equiv a^p +b^p\pmod p ,$$ for $a$ and $b$ indeterminants,
then 
$$(a+b)^{p^n} = \left(\  (a + b )^p\ \right)^{p^{n-1}}\equiv \left(\  a^p + b^p\  \right)^{p^{n-1}}\equiv a^{p^n}+ b^{p^n}\pmod p,$$
which also gives the result.
