# Proving that extension by radicals implies solvable group

I'm trying to understand the following excerpt from Fraleigh's A First Course In Abstract Algebra, Seventh Edition, pp. 472-473:

56.4 Theorem Let $F$ be a field of characteristic zero, and let $F\subseteq E\subseteq K\subseteq \overline F$, where $E$ is a normal extension of $F$ and $K$ is an extension of $F$ by radicals. Then $G(E/F)$ is a solvable group.

Proof. Since $K$ is an extension by radicals, $K = F(\alpha_1,\ldots,\alpha_r)$ where $\alpha_1^{n_1}\in F$ and each $\alpha_i^{n_i}\in F(\alpha_1,\ldots,\alpha_{i-1})$ for $1< i\leq r$. Now form the splitting field $L_1$ of $f_1(x) = x^{n_1} - \alpha_1^{n_1}$ over $F$. Then $L_1$ is a normal extension of $F$, and since $\alpha_1^{n_1}\in F$, the preceding lemma implies that $G(L_1/F)$ is a solvable group. By assumption, $\alpha_2^{n_2}\in F(\alpha_1)\subseteq L_1$, and we form the polynomial $$f_2(x) = \prod_{\sigma\in G(L_1/F)} [x^{n_2} - \sigma(\alpha_2)^{n_2}].$$

My question is, why is Fraleigh able to use the expression $\sigma(\alpha_2)^{n_2}$, when we don't necessarily know that $\alpha_2$ is in the domain of $\sigma$, $L_1$? Shouldn't we be using $\alpha_2^{n_2}$, which we know to be in the domain of $\sigma$, so then the definition of $f_2(x)$ would be $$f_2(x) = \prod_{\sigma\in G(L_1/F)} [x^{n_2} - \sigma(\alpha_2^{n_2})]?$$ I see that if $\sigma(\alpha_2)$ were defined, then by the properties of homomorphisms we would have $\sigma(\alpha_2^{n_2}) = \sigma(\alpha_2)^{n_2}$, but I don't think we know that there is a value for $\sigma(\alpha_2)$. Please advise.