Proof that $[\Bbb{Q}(\sqrt{q_1},\dots,\sqrt{q_r}):\Bbb{Q}]=2^r$

I would like to give a detailed proof of why $[\Bbb{Q}(\sqrt{q_1}, \dots, \sqrt{q_{k+1}}):\Bbb{Q}(\sqrt{q_1},\dots, \sqrt{q_{k}})] = 2$. Then you can see where you have implicitly used what.

Using the inductive hypothesis, $\sqrt{q_k},\sqrt{q_{k+1}},\sqrt{q_{k+1}q_{k}} \notin \Bbb{Q}(\sqrt{q_1},\dots, \sqrt{q_{k-1}})$ (and because $\gcd(q_i, q_j) = 1$).
So suppose $\sqrt{q_{k+1}} \in \Bbb{Q}(\sqrt{q_1},\dots, \sqrt{q_{k}})$.
Then there are $x,y \in \Bbb{Q}(\sqrt{q_1},\dots, \sqrt{q_{k-1}})$ so that $$\sqrt{q_{k+1}} = x + y \cdot \sqrt{q_{k}}$$ $$\text{(inductive hyp. provides } [\Bbb{Q}(\sqrt{q_1},\dots, \sqrt{q_{k}}):\Bbb{Q}(\sqrt{q_1},\dots, \sqrt{q_{k-1}})] = 2) \text{)}$$

(note: $x \neq 0$, because otherwise $q_{k} \mid q_{k+1}$. And also $y \neq 0$, because $\sqrt{q_{k+1}} \notin \Bbb{Q}(\sqrt{q_1},\dots, \sqrt{q_{k-1}})$)

So we get $q_{k+1} = x^2 + y^2q_{k} + 2xy\sqrt{q_{k}}$ and this implies that $$\underbrace{2xy\sqrt{q_{k}}}_{\notin\ L:= \Bbb{Q}(\sqrt{q_1},\dots, \sqrt{q_{k-1}})} = -\underbrace{x^2}_{\in L} - \underbrace{y^2q_{k}}_{\in L} + \underbrace{q_{k+1}}_{\in \Bbb{N} \subset L} $$ which is a contradiction. So $$\sqrt{q_{k+1}} \notin \Bbb{Q}(\sqrt{q_1},\dots, \sqrt{q_{k}}) \implies [\Bbb{Q}(\sqrt{q_1}, \dots, \sqrt{q_{k+1}}):\Bbb{Q}(\sqrt{q_1},\dots, \sqrt{q_{k}})] = 2$$