We want to prove that:
$$\aleph_2^{\aleph_0} = \aleph_2\aleph_1^{\aleph_0}$$
My idea was to approach this by doing a Schroder-Bernstein style argument and proving this by showing two inequalities, but that doesn't seem to work because of the extra $\aleph_2$ on the right side. Any suggestions for this? Thanks.
HINT: Prove that $\aleph_1^{\aleph_0}=2^{\aleph_0}$. Now observe that if $\aleph_2\geq2^{\aleph_0}$ equality ensues; and if the other way around then also equality ensues since in that case $\aleph_0^{\aleph_0}\leq\aleph_2^{\aleph_0}=2^{\aleph_0}$.
More generally, this is Hausdorff's formula for cardinal exponentiation applied for $\alpha=1$.
Edit: Show that generally $\kappa^\lambda=|\{A\subseteq\kappa\mid |A|=\lambda\}|$. Note that there are $\aleph_2$ ordinals of size $\aleph_1$, and every countable subset of $\aleph_2$ is in fact a countable subset of an ordinal of size $\aleph_1$.