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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 $K^{\prime }=K({\zeta }_m)$, which itself is a radical adjunction since ${\zeta }_m^m=1$. The Galois group of $N{K}^{\prime }/K^{\prime }$ 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 $K^{\prime }$. 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 $\sigma \ne 1$, it has an eigenvector $a\ne 0$ with eigenvalue ${\zeta }_p^j\ne 1$. 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. Containment in a radical tower is the valid general conclusion.