Sum of powers in modulo p field

Question

Prove that for all prime numbers $p$, in the field $\mathbb F_{p}$ consistent with modulo $p$ we have $(a+b)^{p}=a^p+b^p$. My first idea is to use the binomial theorem and show that all but the two outer terms equal zero. Any hints why this might be true?

Answer 1

If $0<r<p$, we have that $\binom{p}{r}=\frac{p!}{r!(p-r)!}$ is an integer. How many times does $p$ divide the numerator? The denominator? The fraction as a whole?

Answer 2

You are on the right track. The trick is to show that $p$ divides $\displaystyle \binom{p}{k}$ for $0 < k < p$.

To do this, recall that $\displaystyle \binom{p}{k} = \frac{p!}{(p-k)!k!}$. Rearranging, we get $\displaystyle p! = \Big( (p-k)!k! \Big) \binom{p}{k}$. Notice that $p$ divides the LHS of this equation, so it must also divide the RHS...