I'm trying to understand the proof from Dummit's "Abstract Algebra."
Definition. An element $\alpha$ can be expressed by radicals or solved for in terms of radicals if $\alpha$ is an element of a field $K$ which can be obtained by a succession of simple radical extensions \begin{equation} F=k_0\subsetneq K_1\subsetneq\cdots\subsetneq K_i\subsetneq K_{i+1}\subsetneq\cdots\subsetneq K_s=K \end{equation} where $K_{i+1}=K_i(\sqrt[n_i]{a_i})$ for some $a_i\in K_i$ and $n_i>0$, $i=0,1,\ldots,s-1$. Here $\sqrt[n_i]{a_i}$ denotes some root of the polynomial $x^{n_i}-a_i$. Such a field $K$ will be called a root extension of $F$.
Lemma 14.38. Let $F$ be a field of characteristic $0$. If $\alpha$ is contained in a root extension $K$ as in the definition above, then $\alpha$ is contained in a root extension which is Galois over $F$ and where each extension $K_{i+1}/K_i$ is cyclic.
Proof (with details omitted): There is a Galois root extension $L$ of $F$. Let \begin{equation} F=L_0\subsetneq L_1\subsetneq\cdots\subsetneq L_t=L \end{equation} be a chain of simple radical extensions. We now adjoin to $F$ the $n_i$-th roots of unity for all the roots $\sqrt[n_i]{a_i}$ of the simple radical extensions in the Galois root extension $L/F$, obtaining the field $F'$, say, and then form the composite of $F'$ with the root extension: \begin{equation} F\subseteq F'=F'L_0\subseteq F'L_1\subseteq\cdots\subseteq F'L_t=F'L \end{equation} The field $F'L$ is a Galois extension of $F$ since it is the composite of two Galois extensions. The extension from $F$ to $F'=F'L_0$ can be given as a chain of subfields with each individual extension cyclic (this is true for any abelian extension). Each extension $F'L_{i+1}/F'L_i$ is a simple radical extension and since we now have the appropriate roots of unity in the base fields, each of these individual extensions from $F'$ to $F'L$ is a cyclic extension. Hence $F'L/F$ is a root extension which is Galois over $F$ with cyclic intermediate extensions, completing the proof.
I see that there is a chain of fields \begin{equation} F=F_0\subseteq F_1\subseteq\cdots\subseteq F_u=F' \end{equation} such that each $F_{i+1}/F_i$ is a cyclic extension. But they should be simple radical extensions in order to say that $F'L/F$ is a root extension. How do I prove this? I know the following proposition, but $F_i$ doesn't seem to contain all $n^\text{th}$ roots of unity (which is required in the hypothesis).
Proposition 14.37. Any cyclic extension of degree $n$ over a field $F$ of characteristic not dividing $n$ which contains all the $n^\text{th}$ roots of unity is of the form $F(\sqrt[n]{a})$ for some $a\in F$.