 Solution

Define $\vartheta (x)=x^p$ for every $x\in K$. In characteristic $p$, all intermediate binomial coefficients vanish, so $\vartheta (x+y)=\vartheta (x)+\vartheta (y)$; multiplicativity and preservation of one are immediate. If $x^p=y^p$, then $(x-y{)}^p=0$ and the field property gives $x=y$. Thus this field homomorphism is injective, and finiteness makes it surjective. It is therefore a field automorphism, sending $\lambda$ to ${\lambda }^p$ as required.