TITLE: Prime dividing the binomial coefficients
QUESTION [34 upvotes]: It is quite easy to show that for every prime $p$ and $0<i<p$ we have that $p$ divides the binomial coefficient $\large p\choose i$; one simply notes that in $\large \frac{p!}{i!(p-i)!}$ the numerator is divisible by $p$ whereas the denominator is not (since it is a product of numbers smaller than $p$ and $p$ is prime).
My problem is with generalizing this argument for $q=p^n$. I'm looking for the most elegant and simple way to prove that $p$ still divides $\large q\choose i$.

REPLY [3 votes]: This is a simplification of proof suggested by user84232. (Probably the simplest proof, I see here).
Let $0<i<p^n$. Then $i=p^kj$, where $0\le k<n$, while $p$ and $j$ are co-prime.
Therefore
$$
\binom{p^n}{i}=\frac{p^n}{i}\;\binom{p^n-1}{i-1}= \frac{p^{n-k}}{j}\;\binom{p^n-1}{i-1}
$$
which can be rewritten as 
$$
j\;\binom{p^n}{i}=p^{n-k}\;\binom{p^n-1}{i-1}
$$
Since $j$ and $p^{n-k}$ are co-prime, $\binom{p^n}{i}$ must be divisible by $p^{n-k}$, so (because of $n>k$) divisible by $p$. That's it!