Prove that if $A$ is a countable infinite set, then each Cartesian product $A^n$, with $n \in \mathbb{N}_0$ is also countable infinite.
The book I'm currently using already proved that:
Theorem 1: If $A$ and $B$ are countable infinite sets, then the Cartesian product $A \times B$ is also countable infinite.
So I can use this for proving that $A^n$ is countable infinite.
With help from lulu and especially fleablood, I finally proved this with induction.
Proof
Basic step: $A^1 = A$ which is countable infinite as it is trivial.
Induction step: We assume that $A^k$ is countable infinite, so we need to prove that $A^{k+1}$ is countable infinite.
We know that
$$A^{k+1} = \underbrace{A \times A \times ... \times A}_\text{$k+1$ times}$$ which is equal to $$A^{k+1} = \underbrace{(A \times A \times ... \times A)}_\text{$k$ times} \times A \text{ as } A^{k+1} = A^k \times A$$
By theorem 1 above and the induction step we can say that $A^{k+1}$ is also countable infinite. Therefore, we can say that $A^n$ is countable infinite.