Let $q=p^r,$ where $p\in\mathbb{P}$ is a prime and $r\in\mathbb{N}\setminus\{0\}$ is a natural number (non-zero). How to prove that for each $i\in\{1,2,\ldots,q-1\}$ the binomial coefficient $\binom{q}{i}$ is divisible by $p$?
I find it easy to show that $p|\binom{p}{i},$ but here it's more complicated :/
Since $\binom{p^r}{i}=\frac{p^r!}{i!(p^r-i)!}$, the number of $p$ dividing this number is $$ \sum_{n\ge 1}\left(\left\lfloor\frac{p^r}{p^n}\right\rfloor-\left\lfloor\frac{i}{p^n}\right\rfloor-\left\lfloor\frac{p^r-i}{p^n}\right\rfloor\right). $$ Each term is nonnegative. Hence the sum above is at least $$ \left\lfloor\frac{p^r}{p^r}\right\rfloor-\left\lfloor\frac{i}{p^r}\right\rfloor-\left\lfloor\frac{p^r-i}{p^r}\right\rfloor = 1. $$