Is this proof by induction that $|\Bbb Q^n|=\aleph_0$ correct? Is this proof by induction that $|\Bbb Q^n|=\aleph_0$  correct?
Suppose I know that $|\Bbb Q|=|\Bbb N|=|\Bbb N^2|=\aleph_0\cdot\aleph_0=\aleph_0$.
Proof: 
Suppose $|\Bbb Q^n|=\aleph_0$, then  $|\Bbb Q^{n+1}|=|\Bbb Q^{n}\times\Bbb Q|= |\Bbb Q^{n}|\cdot|\Bbb Q|=\aleph_0\cdot\aleph_0=\aleph_0 {}_{\square}$

Yes, you are correct. In fact, your argument is valid for any set $X$ with $|X| = \aleph_0$, it doesn't have to be $\Bbb Q$. The only possible worries there would be knowing that $|\Bbb Q| = \aleph_0$ and that $\aleph_0 \cdot \aleph_0  = \aleph_0$, but you said you have this results already, though. Good job.
