Proof for: $(a+b)^{p} \equiv a^p + b^p \pmod p$ a, b are integers. p is prime.
I want to prove:
$(a+b)^{p} \equiv a^p + b^p \pmod p$
I know about Fermat's little theorem, but I still can't get it
I know this is valid:
$(a+b)^{p} \equiv a+b \pmod p$
but from there I don't know what to do.
Also I thought about
$(a+b)^{p} = \sum_{k=0}^{p}\binom{p}{k}a^{k}b^{p-k}=\binom{p}{0}b^{p}+\sum_{k=1}^{p-1} \binom{p}{k} a^{k}b^{p-k}+\binom{p}{p}a^{p}=b^{p}+\sum_{k=1}^{p-1}\binom{p}{k}a^{k}b^{p-k}+a^{p}$
Any ideas?
Thanks!

First of all, $a^p \equiv a \pmod p$ and $b^p \equiv b \pmod p$ implies $a^p + b^p \equiv a + b \pmod p$.
Also, $(a+b)^p \equiv a + b \pmod p$.
By transitivity of modulo, combine the above two results and get $(a+b)^p \equiv a^p + b^p \pmod p$.
Done.
