Any Sub-extension of a Radical Extension is Solvable?

Question

I have seen the following theorem stated without proof. (I assume that all fields are characteristic zero.)

Theorem: Let $F\subset K$ be a radical extension of fields. That is, suppose that $K$ can be obtained from $F$ by successive adjunction of radicals. Then for any intermediate field $F\subseteq L\subseteq K$ the group $\mathrm{Aut}_F(L)$ (i.e., automorphisms of $L$ that fix $F$) is solvable.

Can someone please point me to a proof? The application I have in mind is when $L$ is the splitting field of a polynomial.

Answer 1

I found a reference. This is Theorem V.9.4 in Hungerford.