First of all, I am aware that there is another question regarding a proof of the same lemma in this website. The reason as to why I still want to understand the following, different proof is that it is shorter (5 lines of text instead of 20) and that it is in my book.
By definition, $M:K$ is a radical extension if there exists {$x_1, ..., x_r$} so that $M=K(x _1, ..., x_r)$ and $x_i ^{n_i} \in K(x_1, ..., x_ {i-1})$ for some natural number $n_i$.
I understand why there exists a $K$-automorphism $\tau$ of $M$ such that $\tau(\alpha _i) = (\beta_{ij})$, yet I still fail to see how this implies $M:K$ is a radical extension.
I would really appreciate any help/thoughts!
As far as I can see, the relevant facts are these:
Each $\beta_{ij}$ is radical over $K$. This follows from the fact that $\alpha_i$ is radical over $K$, and from the fact that there is an isomorphism $\tau : M \to M$ such that $\tau (\alpha_i) = \beta_{ij}$. Indeed, consider the following inclusions: $$ \tau(\alpha_1)^{n_1} \in K, \ \ \ \ \ \tau(\alpha_2)^{n_2} \in K\left(\tau(\alpha_1)\right), \ \ \ \ \ \dots \ \ \ \ \ \beta_{ij}^{n_i} = \tau(\alpha_i)^{n_i} \in K\left(\tau(\alpha_1), \dots, \tau(\alpha_{n-1}) \right). $$
$M$ is generated over $K$ by all of the $\beta_{ij}$'s. This is simply because $M$ is the splitting field of $ \Pi_i f_i$ over $K$, and the roots of $\Pi_i f_i$ are precisely the $\beta_{ij}$'s.
Having found a set of generators for $M$ over $K$, each of which is radical over $K$, we may conclude that $M$ is radical over $K$.