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How do I prove this theorem?

For a prime number $p$ and integer $i$:

If $0 < i < p$ then $p \mid \binom{p}{i}$.

Alexis Olson
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Miguel
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    This was proved here: https://math.stackexchange.com/questions/328655/proving-prime-p-divides-binompk-for-k-in-1-ldots-p-1 – j4l3kl24jkl2 Nov 04 '16 at 22:25
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    Since ${p \choose i} = \frac {p!}{(p-i)!i!}= p* \frac {(p-1)!}{(p-i)!i!} $and $p$ is prime, that should be pretty obvious, shouldn't it? – fleablood Nov 04 '16 at 22:32
  • @fleablood That requires us to show that $\frac{(p-1)!}{(p-i)!i!}$ is an integer. Of course, it doesn't take more than a sentence or two, but I wouldn't say it's obvious. – Arthur Nov 04 '16 at 22:40
  • @Arthur: Might be, might not be. Yet another reason that it's useful when the OP states their level of understanding. :-) – Brian Tung Nov 04 '16 at 22:41
  • We know ${a \choose b}$ is an integer. And we know that nothing less than $p$ has any factors in common with $p$. No, it isn't obvious. That was me being snide. but looking at factors should be obvious first and seeing where to take it should be clear. – fleablood Nov 04 '16 at 23:01

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Hint: The simplest proof uses the relation $$\binom pk=\frac pk\binom{p-1}{k-1} \quad\text{for all}\; k>0 $$ and Gauss' lemma.

Bernard
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