A similar question was asked before here; but the answer does not quite do it for me.
Theorem (Ian Stewart's Galois Theory 4th edition): a complex number $\alpha$ is an element of the Pythagorean closure $\mathbb{Q}^{py}$ of $\mathbb{Q}$ iff there is a tower of field extensions $\mathbb{Q}=K_0\subseteq K_1\subseteq \dots \subseteq K_n=\mathbb{Q}(\alpha)$ such that $[K_{j+1}: K_j]=2$ for $0\leq j\leq n-1$.
Proof: First, suppose such a tower exists. We prove by induction on $j$ that $K_j \subseteq \mathbb{Q}^{py}$. This is clear for $j=0$. Now, $K_{j+1}$ is an extension of $K_j$ of degree 2, so $K_{j+1}=K_j(\beta)$ where the minimum polynomial of $\beta$ over $K_j$ is quadratic. Since quadratics can be solved by extracting square roots, $\beta \in \mathbb{Q}^{py}$. Therefore $\alpha \in \mathbb{Q}^{py}$.
Next, suppose that $\alpha \in \mathbb{Q}^{py}$. We prove that such a tower exists. By the definition of $\mathbb{Q}^{py}$, there is a tower $$\mathbf{\mathbb{Q}=L_0\subseteq L_1\subseteq\dots \subseteq L_n \supseteq \mathbb{Q}(\alpha)}$$ such that $\mathbf{[L_{j+1}: L_j]=2}$ for $\mathbf{0 \leq j \leq n-1}$.
My question: I don't follow the argument that is highlighted in boldface-it is just too terse. Any help would be greatly appreciated.
Let $F=\{f_1,f_2,f_3,\ldots\}$ be countable and $P(F):=F(\sqrt{1+f_1^2},\sqrt{1+f_2^2},\sqrt{1+f_3^2},\ldots)$.
Then $F^{py}=\bigcup_{n\geq1} P^n(F)$, where $P^n$ is the $n$-fold composition of $P$ with itself.
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For every $k\in K$, the extension $[K(\sqrt{1+k^2}):K]$ is of degree either $2$ or $1$.
Furthermore, each element $\alpha\in P(K)$ can be generated by finitely many such $\sqrt{1+k_i^2}$. (This is what it means to have the smallest extension of $K$ that contains all elements of the form $\sqrt{1+k^2}$ for $k\in K$.)
It's not hard to show by induction on $m$ that each $\alpha\in P^m(K)$ is algebraic over $K$ and can be generated by a tower of degree-$2$ extensions.