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Suppose that $p$ is prime and $0<k<p$. Prove that $p|$${p}\choose{k}$.

I am asked to prove a Lemma beforehand, which follows from Euclid's Lemma. It states: Let $m,n \in\mathbb{Z}$, $p$ be prime, $m|n$, $p|n$, and $p\nmid m$. Then $p|\frac{n}{m}$. After I have done this, I return to the normal problem and try to fulfill the Lemma's assumptions with the current variables.

First, $p|p!=p*(p-1)*...*2*1$ since $p|p$, always. Then, all we have left to show is $1$) $p\nmid k!(p-k)!$ and $2$) $k!(p-k)!|p!$ and apply the Lemma.

I attempt to show $1$) by noting that $p\nmid k!$ and $p\nmid (p-k)!$ since $p$, as a prime, will only divide its multiples which are greater than or equal to $p$. Since both $(p-k)$ and $k$ are strictly less than $p$, they will not contain any multiples of $p$. Then, $p\nmid k!(p-k)!$ by the contrapositive of Euclid's Lemma.

My concerns are these: Is the argument above for $p\nmid k!(p-k)!$ rigorous enough? And, how can I show that $k!(p-k)! | p!$ ?

Did
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  • "And, how can I show that $k!(p-k)! | p!$ ?" Well, $\frac{p!}{k!(p-k)!}={p\choose k}$ is an integer, right? – Did Sep 13 '16 at 21:57
  • Agreed, but @Did, how do you know it is an integer? –  Sep 13 '16 at 21:59
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    The number of putting $k$ objects in $p$ boxes has to be an integer, no? – Paolo Leonetti Sep 13 '16 at 22:00
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    "how do you know it is an integer?" Because Pascal's triangle can be computed recursively line by line, using the identity ${n\choose k}={n-1\choose k}+{n-1\choose k-1}$ and starting from integer valued initial conditions when $n=0$. – Did Sep 13 '16 at 22:03

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