Prove that if $2^{\aleph_3}<\aleph_\omega$, then $\aleph_\omega^{\aleph_3}=\aleph_\omega^{\aleph_0}$

Question

Prove that if $2^{\aleph_3}<\aleph_\omega$, then $\aleph_\omega^{\aleph_3}=\aleph_\omega^{\aleph_0}$

I do not know how to prove it. I need some hint because cardinal arithmetic is something I am very bad at and I have no clear idea which facts should I use here.

Answer 1

To each $f\!:\aleph_3\to\aleph_\omega$ associate the sequence of maps $(f_n)_{n<\omega}$ where $f_n\!:\aleph_3\to\aleph_n$ is given by $f_n(\alpha)=0$ unless $f(\alpha)\in\aleph_n$, in which case $f_n(\alpha)=1+f(\alpha)$. Note that the sequence uniquely identifies $f$.

Now, there are $\aleph_n^{\aleph_3}<\aleph_\omega$ possible values for $f_n$ (this is where the assumption that $2^{\aleph_3}<\aleph_\omega$ is used, via Hausdorff's formula), so there are at most $\aleph_\omega^{\aleph_0}$ such sequences $(f_n)_n$, so $\aleph_\omega^{\aleph_3}\le\aleph_\omega^{\aleph_0}$. The other inequality is clear.