An extension of $\mathbb{Q}$ which contains the $n$-th roots of every element

Question

Consider $\mathbb{Q}$, the field of rational numbers.

Let $K_1\subseteq \mathbb{C}$ be the (minimal) splitting field of the family $\{x^n-a\colon a\in\mathbb{Q}, n\geq 1\}$.

Let $K_2\subseteq \mathbb{C}$ be the splitting field of the family $\{x^n-a\colon a\in K_1, n\geq 1\}$.

Contiinuing this we get an ascending chain $\mathbb{Q}\subseteq K_1\subseteq K_2\subseteq \cdots $ of subfields of $\mathbb{C}$.

Notice that $K_1, K_2,\cdots$ are algebraic extension of $\mathbb{Q}$, hence they are infact subfields of $\overline{\mathbb{Q}}$, the algebraic closure of $\mathbb{Q}$ in $\mathbb{C}$. Thus, $K=\cup_i K_i$ is a subfield of $\overline{\mathbb{Q}}$.

Question: Is $K=\cup_i K_i$ a proper subfield of $\overline{\mathbb{Q}}$.

Answer 1

Every element in this field is solvable by radicals.

So in fact, we can not "write down" an algebraic number not in $K$, but thanks to Evariste Galois, we know it exists.