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Prove that if $p$ is an odd prime and $k$ is an integer satisfying $1\leq k \leq p-1$,then the binomial coefficient
$\binom{p-1}{k} \equiv (-1)^k\pmod p$

I have tried basic things like expanding the left hand side to $\frac{(p-1)(p-2).........(p-k)}{k!}$ but couldn't get far enough.

3 Answers 3

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Hint: $(p-1)(p-2)\cdots(p-k)\equiv(-1)(-2)\cdots(-k)$ because $p\equiv 0$.

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    Got it .Thanks for the hint .2012-06-08
5

Hint: For the first few odd primes $p$, write down the $p$th row of Pascal's Triangle modulo $p$; you'll notice a pattern that should be straightforward to prove. Now use the relation $\binom{p-1}{k-1}+\binom{p-1}{k}=\binom{p}{k}$ and an induction argument.

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    @SaurabhHota: It is complete.2012-06-08
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Recall that we have $(x + 1)^p \equiv x^p + 1 \bmod p$. The ring $\mathbb{F}_p[x]$ is an integral domain, so we can divide and it follows that $(x + 1)^{p-1} \equiv \frac{x^p + 1}{x + 1} \equiv x^{p-1} - x^{p-2} \pm ... - x + 1 \bmod p.$