Let $2\leq q_1<q_2< \dots<q_r$ be square-free natural numbers such that $mcd(q_i,q_j)=1$. Prove that the field extension $\Bbb{Q}(\sqrt{q_1},\dots,\sqrt{q_r})|\Bbb{Q}$ has degree $$[\Bbb{Q}(\sqrt{q_1}, \dots, \sqrt{q_{r}}):\Bbb{Q}]=2^r$$
I tried to prove the assertion by induction over $r$. For $r=1$, we need to prove that $\Bbb{Q}(\sqrt{q})|\Bbb{Q}$ has degree $2$, but this is clear since the minimal polynomial of $\sqrt{q}$ over $Q$ is
$$P(t)=t^2-q$$ which has degree $2$.
Now let's assume the assertion is true for $r=k$ and prove it for $k+1$. Consider
$$\Bbb{Q}(\sqrt{q_1}, \dots, \sqrt{q_{k+1}})=\Bbb{Q}(\sqrt{q_1}, \dots, \sqrt{q_{k}})( \sqrt{q_{k+1}})$$ Then we have the following chain of inclusions
$$\Bbb{Q} \subset \Bbb{Q}(\sqrt{q_1}, \dots, \sqrt{q_{k}}) \subset \Bbb{Q}(\sqrt{q_1}, \dots, \sqrt{q_{k}})( \sqrt{q_{k+1}}) $$
Thus
$$[\Bbb{Q}(\sqrt{q_1}, \dots, \sqrt{q_{k+1}}):\Bbb{Q}]=[\Bbb{Q}(\sqrt{q_1}, \dots, \sqrt{q_{k+1}}):\Bbb{Q}(\sqrt{q_1},\dots, \sqrt{q_{k}})]\cdot [\Bbb{Q}(\sqrt{q_1}, \dots, \sqrt{q_{k}}):\Bbb{Q}]=2\cdot 2^k$$
since the minimal polynomial of $\sqrt{q_{k+1}}$ over $ \Bbb{Q}(\sqrt{q_1}, \dots, \sqrt{q_{r}})$ is
$$P(t)=t^2-q_{k+1}$$
which again has degree $2$.
That should conclude the proof. However, something seems to be wrong, since I didn't use the hypothesis that $mcd(q_i,q_j)=1$ and that each $q_i$ is square-free at all. So I'm afraid that I left something unproved, that requires the use of the hypothesis. I think that might be the part where I stated that the minimal polynomial of $\sqrt{q_{k+1}}$ over the "small field" was the one I said it was, but I really don't know how to prove it. Any help would be appreciated. Thanks in advance!