[step: Direction (⇒): radical extension implies solvable Galois group] Suppose $f$ is solvable by radicals. Then there exists a radical tower $$K = F_0 \subset F_1 \subset \cdots \subset F_m$$ where each $F_{i+1} = F_i(\alpha_i)$ with $\alpha_i^{n_i} \in F_i$ for some positive integer $n_i$, and the splitting field $L$ of $f$ over $K$ satisfies $L \subseteq F_m$. Set $n = n_1 n_2 \cdots n_m$ and let $\zeta$ be a primitive $n$-th root of unity. Form the enlarged tower by adjoining $\zeta$ to every level: $$K(\zeta) = F_0(\zeta) \subseteq F_1(\zeta) \subseteq \cdots \subseteq F_m(\zeta).$$ Each step $F_i(\zeta) \subseteq F_{i+1}(\zeta)$ is still a radical extension: we have $F_{i+1}(\zeta) = F_i(\zeta)(\alpha_i)$ where $\alpha_i^{n_i} \in F_i \subseteq F_i(\zeta)$, and now $F_i(\zeta)$ contains all $n_i$-th roots of unity since $n_i \mid n$. [guided] Why does the presence of $\zeta$ in the base guarantee that each step $F_i(\zeta) \subseteq F_{i+1}(\zeta)$ is a cyclic Galois extension? This is exactly the setting of Kummer theory. Since $F_i(\zeta)$ contains all $n_i$-th roots of unity and $\operatorname{char}(K) = 0$, the extension $F_i(\zeta)(\alpha_i)/F_i(\zeta)$ is Galois with $$\operatorname{Gal}\bigl(F_i(\zeta)(\alpha_i)/F_i(\zeta)\bigr) \hookrightarrow \mathbb{Z}/n_i\mathbb{Z},$$ hence cyclic. This is because every $K$-automorphism $\sigma$ of the upper field must send $\alpha_i$ to $\zeta^{j}\alpha_i$ for some $j$, so $\sigma$ is determined by this single exponent $j \bmod n_i$. [/guided] We also need the bottom step. The extension $K \subseteq K(\zeta)$ is the splitting field of $t^n - 1$, so it is Galois with $$\operatorname{Gal}\bigl(K(\zeta)/K\bigr) \hookrightarrow (\mathbb{Z}/n\mathbb{Z})^{\times},$$ which is abelian. [/step] [step: Assembling the subnormal series and passing to the quotient] Write $E = F_m(\zeta)$. The tower $K \subset K(\zeta) \subset F_1(\zeta) \subset \cdots \subset F_m(\zeta) = E$ is a sequence of Galois extensions where each relative Galois group is abelian (the first step) or cyclic (every subsequent Kummer step). Translating this into a statement about $\operatorname{Gal}(E/K)$: Consider the normal closure $\widetilde{E}$ of $E$ over $K$ (if $E/K$ is not already Galois). In characteristic zero, one can refine the tower so that $\widetilde{E}/K$ is Galois. However, a cleaner route uses the following standard fact: each $F_i(\zeta)/F_{i-1}(\zeta)$ being cyclic Galois means the composite extension $E/K$ has a subnormal series $$\{e\} = \operatorname{Gal}(E/E) \trianglelefteq \operatorname{Gal}(E/F_{m-1}(\zeta)) \trianglelefteq \cdots \trianglelefteq \operatorname{Gal}(E/K(\zeta)) \trianglelefteq \operatorname{Gal}(E/K)$$ where each successive quotient embeds into the corresponding cyclic or abelian group. Hence $\operatorname{Gal}(E/K)$ is solvable. [guided] Now $L \subseteq E$, so how does solvability of $\operatorname{Gal}(E/K)$ give solvability of $\operatorname{Gal}(L/K) = \operatorname{Gal}(f)$? Since $L/K$ is the splitting field of the separable polynomial $f$, it is Galois over $K$. The restriction map $$\operatorname{Gal}(E/K) \longrightarrow \operatorname{Gal}(L/K), \qquad \sigma \longmapsto \sigma\big|_L$$ is a surjective group homomorphism (every automorphism of $L/K$ extends to $E/K$). Its kernel is $\operatorname{Gal}(E/L)$, so $$\operatorname{Gal}(L/K) \cong \operatorname{Gal}(E/K) / \operatorname{Gal}(E/L).$$ A quotient of a solvable group is solvable, so $\operatorname{Gal}(f) = \operatorname{Gal}(L/K)$ is solvable. [/guided] [/step] [step: Direction (⇐): solvable Galois group implies solvable by radicals] Now assume $G = \operatorname{Gal}(L/K)$ is solvable, where $L$ is the splitting field of $f$ over $K$. Since $G$ is a finite solvable group, it admits a subnormal series $$\{e\} = G_r \trianglelefteq G_{r-1} \trianglelefteq \cdots \trianglelefteq G_1 \trianglelefteq G_0 = G$$ with each quotient $G_i / G_{i+1}$ cyclic of prime order $p_i$. Set $n = p_1 p_2 \cdots p_r$ (or more precisely the lcm), let $\zeta$ be a primitive $n$-th root of unity, and form $L' = L(\zeta)$. The extension $K(\zeta)/K$ is radical: $\zeta$ is a root of $t^n - 1 \in K[t]$. Now consider $G' = \operatorname{Gal}(L'/K(\zeta))$. Since $L' = L(\zeta)$ and $K(\zeta)$ is obtained by adjoining roots of unity, we have $G' \hookrightarrow G$ (restriction to $L$ is injective on $\operatorname{Gal}(L'/K(\zeta))$ when $L$ and $K(\zeta)$ are linearly disjoint, and in general $G'$ is isomorphic to a subgroup of $G$). A subgroup of a solvable group is solvable, so $G'$ is solvable and has a subnormal series with cyclic quotients of prime order. [/step] [step: Kummer descent — converting cyclic layers into radical steps] By the Galois correspondence applied to $G' = \operatorname{Gal}(L'/K(\zeta))$ and its subnormal series, we obtain a tower of intermediate fields $$K(\zeta) = K_0 \subset K_1 \subset \cdots \subset K_r = L'$$ where each $K_{i+1}/K_i$ is Galois with $\operatorname{Gal}(K_{i+1}/K_i) \cong G'_i/G'_{i+1}$, a cyclic group of prime order $p_i$. [guided] How does Kummer theory convert each cyclic Galois step $K_{i+1}/K_i$ of prime degree $p_i$ into an adjunction of a radical? Since $K_i \supseteq K(\zeta)$ contains all $p_i$-th roots of unity (as $p_i \mid n$) and $\operatorname{char}(K) = 0$, the Kummer theory correspondence applies. It states: every cyclic Galois extension of degree $p_i$ over a field containing the $p_i$-th roots of unity has the form $K_i(a_i^{1/p_i})$ for some $a_i \in K_i$. Explicitly, there exists $a_i \in K_i^{\times}$ such that $$K_{i+1} = K_i\bigl(a_i^{1/p_i}\bigr), \qquad (a_i^{1/p_i})^{p_i} = a_i \in K_i.$$ This is precisely a radical step. [/guided] Concatenating all these steps, we obtain the radical tower $$K \subset K(\zeta) \subset K_1 \subset K_2 \subset \cdots \subset K_r = L' \supseteq L.$$ The first step adjoins a root of $t^n - 1$ (a radical step with exponent $n$), and each subsequent step $K_i \subset K_{i+1}$ adjoins a $p_i$-th root of an element of $K_i$. Since $L \subseteq L' = K_r$ and $K_r$ is obtained from $K$ by a radical tower, the polynomial $f$ is solvable by radicals. [/step] [step: Conclusion] Combining both directions: $(\Rightarrow)$ If $f$ is solvable by radicals, then its splitting field $L$ embeds into a radical tower. Adjoining enough roots of unity converts each radical step into a cyclic Kummer extension. The resulting Galois group $\operatorname{Gal}(E/K)$ has a subnormal series with abelian quotients, hence is solvable. Since $\operatorname{Gal}(L/K)$ is a quotient of this solvable group, it too is solvable. $(\Leftarrow)$ If $\operatorname{Gal}(f)$ is solvable, its subnormal series with cyclic quotients, combined with the adjunction of appropriate roots of unity, allows Kummer theory to realize each cyclic layer as a radical extension. The resulting tower is a radical extension containing the splitting field of $f$. Therefore, for $K$ of characteristic zero and $f \in K[t]$ separable: \begin{align*} f \text{ is solvable by radicals} \quad\Longleftrightarrow\quad \operatorname{Gal}(f) \text{ is solvable.} \end{align*} $\blacksquare$ [/step]