Is this proof by induction that $|\Bbb Q^n|=\aleph_0$ correct?

Question

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 $|\Bbb Q^{n+1}|=|\Bbb Q^{n}\times\Bbb Q|= |\Bbb Q^{n}|\cdot|\Bbb Q|=\aleph_0\cdot\aleph_0=\aleph_0 {}_{\square}$

Answer 1

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.