# Does the theorem work for even primes (i.e. p=2)

In a question, it was asked to prove that if $p$ is an odd prime, $n>0$ and $0<k<p^n$, 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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Why don't you show your proof? –  draks ... Aug 31 '12 at 6:24
@draks, how about $\binom{p^n}{k} = \frac{p^n}{k} \binom{p^n-1}{k-1} \equiv p^n \cdot k^{-1} \binom{p^n-1}{k-1} \equiv 0 \pmod{p^n}$, which works as $k$ is invertible $\pmod{p^n}$. –  Karolis Juodelė Aug 31 '12 at 6:32
@KarolisJuodelė Thanks! That is essentially how I proved it. $(k,p)=1 \Rightarrow (k,p^n)=1$ so by Euclid's lemma, $p^n \mid k {p^n \choose k}$ and $(k,p^n)=1$ implies $p^n \mid {p^n \choose k}$ –  yoyostein Aug 31 '12 at 6:37
@KarolisJuodelė: why not post that as an answer? –  JavaMan Aug 31 '12 at 7:16
@KarolisJuodelė What is meant by $k$ is invertible $\pmod{p^n}$?Can you explain the last step? –  Saurabh Aug 31 '12 at 7:24

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.