I'm trying to compute the cardinality of a determined set. $A=\{f\colon \omega_n \to \omega_m | |\operatorname{Supp}(f)|=\aleph_k \quad k<n\}.$
As suggested by the exercise, I first tried few elementar cases: $A_1=\{f\colon \omega \to \omega | |\operatorname{Supp}|<\aleph_0\}$.
It's easy to see that $|A_1|=\aleph_0$ as it is a countable union of countable sets.
Another case:
$A_2=\{f\colon \omega_1 \to \omega | |\operatorname{Supp}(f)|=\aleph_0\}$. For this I started having few difficulties, but I think I figured out the solution: $\phi:A_2 \to [\omega_1]^{\aleph_0} \times Fun(Supp(f),\omega) \\ f \mapsto (Supp(f), f|_{Supp(f)})$, is a bijection so we have that $|A_2|=|[\omega_1]^{\aleph_0} \times Fun(Supp(f),\omega)|=2^{\aleph_0}$
So, by using the same argument i want to say that $$\phi:A_2 \to [\omega_n]^{\aleph_k} \times Fun(\operatorname{Supp}(f),\omega_m) \\ f \mapsto (Supp(f), f|_{Supp(f)})$$ is a bijection so $|A|=|[\omega_n]^{\aleph_k} \times Fun(\operatorname{Supp},\omega_m)| = max(2^{\aleph_k},\aleph_n, \aleph_m^{\aleph_k})$.
Is my proof true? Thanks in advance.