Proof. Let $\overline{F}$ denote an algebraic closure of $F$ Assume for the sake of contradiction that $\sqrt[n]{x}\in\overline{F}$ Then since algebraic numbers are closed under multiplication (and thus exponentiation by positive integers), we have $(\sqrt[n]{x})^n=x\in \overline{F}$ so that $x$ is algebraic over $F$ creating a contradiction.