= Solution
The <binomial theorem> isolates the only terms that can survive reduction modulo the <prime number> $p$:
$$
(a+b)^p=a^p+b^p+\sum_{r=1}^{p-1}\binom pr a^{p-r}b^r.
$$
For $1\le r\le p-1$, neither $r!$ nor $(p-r)!$ is divisible by $p$, whereas $p!$ is. Consequently $p\mid\binom pr$, and every intermediate term vanishes in <modular arithmetic>. Thus
$$
\boxed{(a+b)^p\equiv a^p+b^p\pmod p.}
$$
This works for arbitrary <integers> $a,b$, including those divisible by $p$. It is the additive identity behind the <Frobenius endomorphism> in <characteristic> $p$.
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