I am trying to prove the following: if $a,b \in \mathbb{Z}$ then $(a+b)^p \equiv a^p + b^p$ (mod $p$), where $p$ is prime. I am recommended to use the fact that $p \choose k$ is divisible by p for $1 \leq k < p$.
I am trying to set it up so that $(a+b)^p - (a^p + b^p) \equiv 0$ (mod $p$) by noticing that:
$(a+b)^p = \sum_{k=0}^{p}$ $p \choose k$ $a^{p-k} \ b^{k} = pr $ $\sum_{k=0}^{p} a^{p-k} \ b^{k}$, where $r \in \mathbb{Z}$. However, unless I'm missing something, this doesn't seem like the best approach since I won't be able to isolate the first and last terms of this summation (which is what I'm looking for).
Any help is appreciated. Thanks in advance.
hint: $\binom{p}{0} = \binom{p}{p} = 1$