Given that $K$ is a field, $\text{char}(K)=p$ (where $p$ is prime) we need to show that for any integer $n$ the equality
$$ (a+b)^{p^n} =a^{p^n}+b^{p^n} $$ We have the following
$$n=0:\qquad(a+b)^{p^0}=a+b=a^{p^0}+b^{p^0}$$ $$\text{Induction hypothesis}:\qquad (a+b)^{p^n}=a^{p^n}+b^{p^n}$$ $$n+1 \text{ case}:\qquad (a+b)^{p^{n+1}}=((a+b)^{p^{n}})^p=(a^{p^n}+b^{p^n})^p$$ $$=\sum_{k=0}^{p} \binom{p}{k}(a^{p^n})^{p-k}(b^{p^n})^{k}$$ $$=a^{p+1}+b^{p+1}+\sum_{k=1}^{p-1} \binom{p}{k}(a^{p^n})^{p-k}(b^{p^n})^{k}$$
In order to complete the proof we need to get $\sum_{k=1}^{p-1} \binom{p}{k}(a^{p^n})^{p-k}(b^{p^n})^{k}=0$ and we know that we need to use the characteristic property of $K$ somehow, however we don't know in what way.
Write $\displaystyle \binom{p}{k} = \frac{p\times(p-1)\times\cdots\times(p-k+1)}{1\times 2\times\cdots\times k}$, recall that $\displaystyle \binom{p}{k}$ is an integer, and argue that none of those multiplicands in the denominator can cancel out that $p$ in the numerator: they must be cancelling out with the $(p-1)\times\cdots\times(p-k+1)$ part to leave you with an integer that is obviously a multiple of $p$.