*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!