In characteristic $0$, a polynomial solvable by radicals has a solvable Galois group

By [F1], choose a finite radical extension $L\mathrm{/}F$ with $E\subseteq L$. Replacing $L$ by its normal closure over $F$, [L1] lets us assume from the start that $L\mathrm{/}F$ is finite Galois and radical.

Let the radical tower for $L\mathrm{/}F$ use exponents $m_1,\dots ,m_r$, and let $N=m_1\cdots m_r$. Adjoin ${\mu }_N$ to $L$. By [L2], solvability of $\mathrm{Gal}(L\mathrm{/}F)$ is equivalent to solvability of $\mathrm{Gal}(L({\mu }_N)\mathrm{/}F)$. So it is enough to prove the latter solvable.

After adjoining ${\mu }_N$, every step of the radical tower becomes a finite Galois extension with cyclic Galois group: if $F_i=F_{i-1}({\alpha }_i)$ with ${\alpha }_i^{m_i}\in F_{i-1}$, then the enlarged lower field already contains ${\mu }_{m_i}$, so every root of $x^{m_i}-{\alpha }_i^{m_i}$ is $\zeta {\alpha }_i$ with $\zeta \in {\mu }_{m_i}$. Thus $F_i({\mu }_N)\mathrm{/}{F}_{i-1}({\mu }_N)$ is the splitting field of a separable polynomial, and every automorphism is determined by ${\alpha }_i\mapsto \zeta {\alpha }_i$, so its Galois group embeds in the cyclic group ${\mu }_{m_i}$. In particular each step has solvable Galois group, and repeated use of [L3] up the tower makes $\mathrm{Gal}(L({\mu }_N)\mathrm{/}F({\mu }_N))$ solvable.

The cyclotomic extension $F({\mu }_N)\mathrm{/}F$ has abelian Galois group, hence solvable. Applying [L3] once more to the tower $F\subseteq F({\mu }_N)\subseteq L({\mu }_N)$ shows that $\mathrm{Gal}(L({\mu }_N)\mathrm{/}F)$ is solvable. By step 2.1 the same is true of $\mathrm{Gal}(L\mathrm{/}F)$.

The splitting field $E$ is an intermediate field of the finite Galois extension $L\mathrm{/}F$, so $\mathrm{Gal}(E\mathrm{/}F)$ is a quotient of a subgroup of $\mathrm{Gal}(L\mathrm{/}F)$. Therefore [L3] makes $\mathrm{Gal}(E\mathrm{/}F)$ solvable.