If $p$ is a prime number greater than 2 and $k\in \mathbb{N}$ so that $k < p$, how can I prove that $p\choose k$ is congruent to $0 \bmod p$?
If $p$ is a prime number greater than $2$ and $k$ is a natural number so that $k, how can I prove that?
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binomial-coefficients
modular-arithmetic
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1Note natural means $k \geq 1.$ This has probably been asked several times before, it's a standard and useful fact. – 2012-12-05