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In a question, it was asked to prove that if $p$ is an odd prime, $n>0$ and $0, and $\gcd(k,p)=1$, then

${p^n\choose k}\equiv 0 \pmod {p^n}$

My question is, is the hypothesis $p$ is an odd prime necessary? I have worked out a proof, but without using the fact that $p$ is odd, ie., my proof seems to work perfectly for $p=2$.

Sincere thanks for help.

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    @KarolisJuodelė: I'm aware that you didn't write anything OP didn't know, but now it looks like the question has not been answered. Plus, why not copy your comments into an answer and then get credit for your response? (see http://meta.math.stackexchange.com/questions/1559/dealing-with-answers-in-comments for more). – 2012-08-31

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The statement does indeed hold for $2$ as well as any other prime. Both proofs given in comments are correct and don't depend on $p$ being odd.