Proof. Let $\zeta_n$ be a primitive $n^{\mathrm{th}}$ root of unity, and define $F'=F(\zeta_n)$ $L'=L(\zeta_n)$ and $K'=K(\zeta_n)=F'(\sqrt[n]\alpha)$ $$\xymatrix @R1pc@C.3pc{ & K'=K(\zeta_n)=F'(\sqrt[n]\alpha) \ar@{-}[dl]\ar@{-}[dr] & & \\ K=F(\sqrt[n]\alpha) \ar@{-}[dr] & & L'=L(\zeta_n) \ar@{-}[dl]\ar@{-}[dr] & \\ & L \ar@{-}[dr] & & F'=F(\zeta_n) \ar@{-}[dl] \\ & & F & } $$ Now, $L'/F'$ is Galois since $L/F$ is. But $K'$ is a Kummer extension of $F'$ so has cyclic Galois group and thus $L'/F'$ has cyclic Galois group as well (being a quotient of $\Gal(K'/F')$ . Thus $L'$ is a Kummer extension of $F'$ so that $L'=F'(\sqrt[n]{\beta})$ for some $\beta\in L$ It follows that $L=F(\sqrt[n]{\beta})$ But since $L$ is Galois over $F$ it follows that $n\leq 2$ (since otherwise in order to be Galois, $L$ would have to contain the non-real $n^{\mathrm{th}}$ roots of unity).