Let $k$ be an algebraically closed field of characteristic $p>0$ and $K/k$ be a function field, i.e. $K$ is finite over $k(t)$. Consider the field extension $K \subseteq K^{1/p}$. Why does it have degree $p$?
In the case $K = k(t)$ this is clear, because then $K^{1/p} = k(t^{1/p})$ and $t^{1/p}$ has degree $p$ over $K$, since $x^p - t$ is irreducible over $K$ and has $t^{1/p}$ as a root. But I don't know how to deal with the general case.
Remark: The equality $[K^{1/p} : K] = p$ is used in Hartshorne's book in the context of the Frobenius morphism of curves.