Prime-power binomial divisibility
= Prime-power binomial divisibility
{title2=$\binom{p^r}{j}\equiv0\pmod p\quad(0<j<p^r)$}
For a <prime number> $p$ and integer $r\ge1$, every interior <binomial coefficient> in row $p^r$ is divisible by $p$. The polynomial identity $(1+x)^p\equiv1+x^p\pmod p$ iterates under the <Frobenius endomorphism> to $(1+x)^{p^r}\equiv1+x^{p^r}\pmod p$. This concerns divisibility by $p$, not necessarily by $p^r$.