= Solution
For $1\le j\le p-1$, the identity for <binomial coefficients>
$$
j\binom pj=p\binom{p-1}{j-1}
$$
shows that $p$ divides $j\binom pj$. Since $p$ is a <prime number> and $j$ is not divisible by $p$, cancellation modulo $p$ gives
$$
\boxed{\binom pj\equiv0\pmod p\quad(0<j<p).}
$$
The <binomial theorem> consequently gives $(1+x)^p\equiv1+x^p$ as a coefficientwise polynomial <modular congruence>. Apply the same identity with $x$ replaced by $x^{p^r}$ and use <mathematical induction> to obtain
$$
(1+x)^{p^i}\equiv1+x^{p^i}\pmod p\qquad(i\ge1).
$$
This is an iteration of the <Frobenius endomorphism> in characteristic $p$. Comparing the intermediate coefficients gives the <prime-power binomial divisibility> result
$$
\boxed{\binom{p^i}{j}\equiv0\pmod p\quad(0<j<p^i).}
$$
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