Proposition 37 on page 549 of Abstract Algebra, 3rd Ed. by Dummit and Foote claims that irreducible implies separable over a finite field.
Suppose $p(x)$ is irreducible over a finite field of characteristic $p$ and (for the purpose of arriving at a contradiction) suppose $p(x)$ is not separable.
From this I can only conclude that $p'(x)$ is zero for at least one value. However, the authors make the claim that the derivative is identically zero (equivalently, each element in the domain is a multiple root) by claiming that $p(x)$ is a polynomial in $x^p$.
Why is $p(x)$ necessarily a polynomial in $x^p$? And why is it the case that over a finite field we have that either $p'(x)$ is never zero or is identically $0$?