= Solution
Here one must distinguish two conventions for a <radical extension>. The solvability theorem says that a finite extension with solvable normal-closure <Galois group> is contained in a tower of radical adjunctions. Under the convention that this containment defines a radical extension, this is the requested assertion.
Let $N/K$ be the finite normal closure and let $m$ be divisible by the orders of its <Galois group> elements. Adjoin a primitive $m$th <root of unity>, obtaining $K'=K(\zeta_m)$, which itself is a radical adjunction since $\zeta_m^m=1$. The <Galois group> of $NK'/K'$ embeds in the solvable group of $N/K$, so it is solvable. A composition series with cyclic prime-order quotients gives a tower of cyclic prime-degree extensions above $K'$. Each is a radical adjunction, as the following argument shows.
For a cyclic extension $E/F$ of prime degree $p$ containing $\zeta_p$, let $\sigma$ generate its <Galois group>. The $F$-linear operator $\sigma$ satisfies $X^p-1$, a product of distinct linear factors in characteristic zero. Thus it is diagonalizable. Because $\sigma\ne1$, it has an eigenvector $a\ne0$ with eigenvalue $\zeta_p^j\ne1$. Then $a^p\in F$ and $a\notin F$, so $E=F(a)$ by primeness of the degree. Applying this at every step produces a radical tower containing $L$.
If “radical extension” instead means that $L/K$ itself must be such a tower, the printed statement is false without this qualification. For example the real cubic subfield of $\mathbb Q(\zeta_7)$ is cyclic and solvable, but contains no nonrational $a$ with $a^n\in\mathbb Q$: all three conjugates are real roots of the same equation $X^n=a^n$, which has at most two real roots, while any nonrational element of the cubic field has three distinct conjugates. Consequently no nontrivial radical adjunction can start a tower inside it. \b[Containment in a radical tower is the valid general conclusion.]
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