Countably Infinite Cartesian Product I'm having an extremely hard time figuring out how to prove this, would you have to start from one side of the proof and move to the other or prove with induction?
For any $n\in\mathbb{N}$, prove that the cartesian product $A^n = A\times,\dotsc,\times A$ ($n$ times) is countably infinite if $A$ is countably infinite.
Thanks so much for the help!
 A: You want to show that $A$ countable $\implies A^n$ countable. If $A$ is countable there is $f : A \to \mathbb N$ injective. We can create a new injection $g : A^n \to \mathbb N^n$ for any $n$, defined $$(a_1, a_2, \dots, a_n) \mapsto (f(a_1), f(a_2), \dots, f(a_n)).$$ To show $A^n$ countable it will suffice to find an injection $h : \mathbb N^n \to \mathbb N$; then the composition $h \circ g$ will give the desired map $A^n \to \mathbb N$.
The famous injective example $\mathbb N \times \mathbb N \to \mathbb N$ is sometimes called the Cantor pairing function. Rather than duplicating the explanation here I'd refer you to this Math Stack Exchange thread and this Wikipedia article. The idea is to represent $\mathbb N \times \mathbb N$ as a grid, and count the elements, starting at $(0,0)$ and weaving along. (This is the basis for showing $\mathbb Q$ and $\mathbb N$ are of the same cardinality.)
Now, if $k : \mathbb N\times \mathbb N \to \mathbb N$ is our injection, consider $\hat k : (\mathbb N \times \mathbb N) \times \mathbb N \to \mathbb N$ defined 
$((n,m),l) \mapsto k(k(n,m),l)$. The argument extends inductively.
A: Your question is basically asking if the equality $|\Bbb N|^n=|\Bbb N|$ holds.
This is true, using the lemma below we have that 
$$|\Bbb N^{n+1}|=|\Bbb N^{n}\times\Bbb N|=|\Bbb N^n|\cdot |\Bbb N|=|\Bbb N|\cdot|\Bbb N|=|\Bbb N|_\square$$

Lemma. $|\Bbb N\times \Bbb N|=|\Bbb N|$
Let $f:\Bbb N \to \Bbb N^2, x\mapsto (x,0)$. This is injective, thus $|\Bbb N|\le |\Bbb N^2|$.
Let $g:\Bbb N^2 \to \Bbb N, (x,y)\mapsto 2^x3^y$, if this is injective, we're done, is it?
