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It is easy to prove that $p\mid \binom pk$ for $k=1,2,\ldots,p-1$, so that for $x,y\in R$ a commutative ring of characteristic $p$ a prime, $$(x+y)^p = \sum_{k=0}^p \binom pk x^ky^{p-k} = x^p + y^p. $$ But is there reason a priori that the $\binom pk$ should be multiples of $p$?

Bernard
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Math1000
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    There are several approaches you could look at, listed here -- https://math.stackexchange.com/questions/328655/proving-prime-p-divides-binompk-for-k-in-1-ldots-p-1 – PrincessEev Dec 28 '19 at 10:10
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    What do you mean with an a priori reason? – Bernard Dec 28 '19 at 10:10
  • What I mean by "a priori" is an intuition as to why the $\binom pk$ should be multiples of $p$, not a detailed proof. – Math1000 Dec 28 '19 at 10:30
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    If $k\ge 1$, then $p$ occurs in the numerator of $\dfrac{p!}{k!(p-k)!}$ but not the denominator. That should be a priori enough for anybody. – TonyK Dec 28 '19 at 10:32

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I am just rewriting/rephasing TonyK's comments. By definition, for $k>1$we have that $$\binom pk = \dfrac{p!}{k!(p-k)!} = \frac{1.2.3...p}{(1.2.3...k)(1.2.3...(p-k))}.$$ Any factor at the bottom is clearly less than $p$ and since $p$ is a prime none of the factors of the bottom will divide $p$. Therefore, $\binom pk$ must be a multiple of $p$. I think the whole story is simply the consequence of the primeness of $p$. I convince my self by the following identity: $$ \binom pk. k!= ^pP_k .$$ The right side is clearly a multiple of $p$, and because of the primeness of $p$, we cannot cancel this $p$ guy. I know there is nothing important here. Just rephasing the same thing. So, this should not be a good answer to your question. Finally, I don't see any interesting reason that forces $\binom nk$ to be a multiple of $p$ besides the primeness of $p$.

user26857
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Matha Mota
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