All the binomial coefficients $\binom{n}{ i}$ are divisible by a prime $p$ only if $n$ is a power of $p$.

Question

I'm looking for a "high school / undergraduate" demonstration for the:

All the binomial coefficients $\binom{n}{i}=\frac{n!}{i!\cdot (n-i)!}$ for all i, $0\lt i \lt n$, are divisible by a prime $p$ only if $n$ is a power of $p.$

There is a "elementary" (good for high school) demonstration of reciprocal here: http://mathhelpforum.com/number-theory/186439-binomial-coefficient.html

And for this part I'm trying to avoid using more advanced tools like the Theorem of Lucas or Kummer.

Thanks in advance to anyone who can help.

Answer 1

Let $a=p^nk$. Then let $p^m$ be the greatest power of $p$ smaller than or equal to $p^nk$. This number must be strictly smaller than $p^nk$.

Look at $$\binom{p^nk}{p^m}=\frac{ p^nk\cdot(p^nk-1)\cdot(p^nk-2)\dots(p^nk-(p^nk-p^m+1)}{p^m\cdot (p^m-1)\dots 1}$$ The factors on the top form a complete residue class mod $p^m$, as do the elements in the bottom. Therefore we can pair them up so they are congruent mod $p^m$. But since no number less than or equal to $p^nk$ is divisible by $p^ {m+1}$ this gives us all the information to determine the maximum power of $p$ that divides both the top and the bottom, and it is the same. Therefore that fraction is not divisible by $p$.