Let $p$ be a prime, we want to prove that $(x+y)^p=x^p+y^p \mod p$. The left handside looks like something we can throw the binomial theorem at, also in the previous proof the book asked to prove that $p|\binom{p}{k}$ for all natural numbers $k$ except $0$ and $p$. So we then know that $\binom{p}{k} \equiv 0 \mod p$ for k unequal to $0$ and to p$.
$$(x+y)^p=\sum^p_{k=0} \binom{p}{k} x^{p-k} y^k \equiv \binom{p}{p}x^{p-p}y^p+ \binom{p}{0}x^{p-0}y^0 \mod p $$ Since all other terms are congruent to $0$. This can be simplified to:
$$(x+y)^p=x^p+y^p\mod p $$
Alternatively, for a way shorter proof we can just note that for any prime $p$ we have that $a^p \equiv a \mod p$ by Fermat's little theorem. $$ (x+y)^p \equiv x+y \mod p \equiv x^p +y^p \mod p$$
Is this correct?