Suppose we want to show that a Cartesian product of the form $B^n$ of $n$ copies of the natural numbers is countably infinite. The proof with just two copies of the natural numbers is a standard Cantor diagonalization argument, but I don't understand the inductive step of this proof, which seems to be that $B^{k+1} = B^k \times B$, and since the Cartesian product of two copies of the naturals is countably infinite, this Cartesian product of two countably infinite sets is also countable.
This doesn't seem very convincing. Wouldn't we need an additional lemma to say that the Cartesian product of two countably infinite sets, regardless of how many items are accounted for in their Cartesian product, is also countably infinite? Or do we create a bijection from $B^k$ to $\mathbb{N}$ and $B$ to $\mathbb{N}$ and then show countability by function composition? This doesn't seem like a very difficult proof, but it seems that there's a jump in the logic.
This is an answer to your last question.
The Cartesian product of two sets is a set of couples $$A\times B=\{(a,b)\ | a\in A, b\in B\}$$
The Cartesian product of three sets is a set of triples defined similarly $$A\times B\times C=\{(a,b, c)\ | a\in A, b\in B, c\in C\}$$
Therefore there is a slight conceptual difference between, say $(A\times B)\times C$ and $A\times B\times C$: an element in the former will look like $$((a, b), c)$$ which is really a couple (whose first element is a couple itself) and not equal to the triple $(a,b,c)$.
Yet there is a natural identification between the two sets (which is even an isomorphism for most of the classical algebraic structures) and in general we omit to mention it since it is harmless for most purposes.