How to prove that this infinite extension of $\mathbb{Q}$ is Galois Let $K_0=\mathbb{Q}$ and for $n>0$ define $K_{n+1}$ as the extension of $K_n$ obtained by adjoining to $K_n$ all the radicals of elements in $K_n$. Let $K$ be the union of the subfields $K_n$.
Prove that $K$ is a Galois extension of $\mathbb{Q}$.
By definition, $K_{n+1}/K_n$ is algebraic, hence separable ($char\mathbb{Q}=0 )$
I do not know how to prove the normality.

We have to show that every homomorphism $\phi : K \to \overline{\mathbb{Q}}$ factors through $K$ (this is equivalent to normality). In fact we can prove $\phi(K_n) \subseteq K_n$ by induction on $n$. For $n=0$ this is clear, and for $n \leadsto n+1$ let $a \in K_{n+1}$, wlog a radical from some element $K_n$, i.e. there is some $d \geq 1$ with $a^d \in K_n$. It follows $\phi(a)^d = \phi(a^d) \in \phi(K_n) \subseteq K_n$, and therefore $\phi(a) \in K_{n+1}$.
By the way, the main result of Galois theory implies that $K$ is the compositum of all finite solvable Galois extensions of $\mathbb{Q}$ (i.e. those with solvable Galois group), sometimes denoted by $\mathbb{Q}^{\mathrm{solv}}$. See MO/4379 for some information about it. It follows that $\mathrm{Gal}(\mathbb{Q}^\mathrm{solv}/\mathbb{Q})$ is the maximal pro-solvable quotient of $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$. I wonder if this profinite group  has been classified?
