
One more proof
Prove that for all natural numbers k,m, Nk X Nm is finite.
I know the lemma "for every k belong to N, every subset of Nk is finite", could we just say that Nk X Nm is a subset of Nk, so it is finite....?
The hint is "consider the function f: NkXNm>Nkm given by f(a,b)=(a1)m+b. Use the Division Algorithm to show f is a onetoone correspondence. This is a formal version of the product rule that says the number of elements in NkXNm is km
This makes it more complicated than I thought.

I suggest this mapping: $\displaystyle \Phi :N_k \times N_m \mapsto N_{\left( {2^k } \right)\left( {3^m } \right)} \,,\,\Phi (g,h) = \left( {2^g } \right)\left( {3^h } \right)$.
Of course $\displaystyle N_{\left( {2^k } \right)\left( {3^m } \right)} \,$ is finite.
It is easy to show that $\displaystyle \Phi$ is one to one.
Here are theorems that you need to do these sorts of cardinality problems.
T1 There is an injection from A to B if and only if there is a surjection from B to A.
T2 If there is an injection from A to B then the cardinality of A at most the cardinality of B.
i.e. $\displaystyle A\le B$.
T3 If there is an surjection from B to A then the cardinality of B at least the cardinality of A.
i.e. $\displaystyle B\ge A$.