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Hi. I don't understand the following proof.

Let p be a prime number and 1<=k<=p-1, then:

[itex]{p \choose k} \equiv 0 \pmod{p}[/itex]

[itex]{p \choose k} = \frac{p!}{k!(p-k)!}[/itex]. Since [itex]p | p![/itex] but [itex] p \not | k! \land p \not | (p-k)![/itex] the proof follows.

I think here what you want to show is that p divides the binomial coefficient, but how does the proof do that?

**Lemma**Let p be a prime number and 1<=k<=p-1, then:

[itex]{p \choose k} \equiv 0 \pmod{p}[/itex]

**Proof**[itex]{p \choose k} = \frac{p!}{k!(p-k)!}[/itex]. Since [itex]p | p![/itex] but [itex] p \not | k! \land p \not | (p-k)![/itex] the proof follows.

I think here what you want to show is that p divides the binomial coefficient, but how does the proof do that?

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