Corollary 1

— Every algebraic extension of a perfect field is a perfect field.

Let $K$ be a perfect field of characteristic exponent $p$, and let $E$ be an algebraic extension of $K$. Then $E$ is separable over $K$ ($V$, p. 36, Prop. 2) whence $E = K(E^p)$ by Prop. 11; but we have $K = K^p \subset E^p$, hence $E = K(E^p) = E^p$, and so $E$ is perfect.