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
$(a+b)^{p} = \sum_{k=0}^{p}\binom{p}{k}a^{k}b^{p-k}=\binom{p}{0}b^{p}+\sum_{k=1}^{p-1}a^{k}b^{p-k}+\binom{p}{p}a^{p}=b^{p}+\sum_{k=1}^{p-1}\binom{p}{k}a^{k}b^{p-k}+a^{p}$
Any ideas?


Solutions Collecting From Web of "Proof for: $(a+b)^{p} \equiv a^p + b^p \pmod p$"

Your second idea is good, so let’s work a little bit on it: We have that $(a+b)^p=a^p+b^p+\sum\limits_{k=1}^{p-1}{p\choose k}a^{k}b^{p-k}$. Obviously it is enough to show that each term of this sum is divisible by $p$ in order to get that the whole sum is $\equiv 0\mod p$.

So why is that the case? For $1\leq k\leq p-1$ we have that ${p\choose k}=\frac{p\cdot (p-1)!}{k!(p-k)!}$ and since $p$ is prime, no factor in the denominator divides $p$, so the denominator does not divide $p$ at all: Hence we have that already $\frac{(p-1)!}{k!(p-k)!}$ is integer and so $p\mid{p\choose k}$. Of course then ${p\choose k}a^kb^{p-k}$ is divisible by $p$ and hence the whole sum is too.