 Prime-power binomial divisibility

For a prime number $p$ and integer $r\ge 1$, every interior binomial coefficient in row $p^r$ is divisible by $p$. The polynomial identity $(1+x{)}^p\equiv 1+x^p(\mathrm{mod}p)$ iterates under the Frobenius endomorphism to $(1+x{)}^{p^r}\equiv 1+x^{p^r}(\mathrm{mod}p)$. This concerns divisibility by $p$, not necessarily by $p^r$.