TITLE: Rank $ (a) {\aleph_4}^{\aleph_4}, (b) {\aleph_0}^{\aleph_4}, (c) {\aleph_4}^{\aleph_0}, (d) {\aleph_4}, (e) {\aleph_5}, (f) {\aleph_5}^{\aleph_0} $
QUESTION [6 upvotes]: If we work in a model of set theory where $2^{\aleph_0} = \aleph_5$, then Rank in ascending order the following cardinals:
$$
(a) {\aleph_4}^{\aleph_4}, (b) {\aleph_0}^{\aleph_4}, (c) {\aleph_4}^{\aleph_0}, (d) {\aleph_4}, (e) {\aleph_5}, (f) {\aleph_5}^{\aleph_0}
$$
If possible, specify whether there is a strict equality or inequality
My attempt:
the only one I could figure out was: (f) $ {\aleph_5}^{\aleph_0} = (2^{\aleph_0})^{\aleph_0} = 2^{\aleph_0} $.
What will be $\aleph_4$ here?

REPLY [2 votes]: $\aleph_4$ is simiply the fourth smallest infinite (well-ordered, if you work in ZF) cardinal.

From what you wrote, $\aleph_5=\aleph_5^{\aleph_0}$, so you have one equality.
By definition, $\aleph_4<\aleph_5$.
Clearly, $\aleph_4^{\aleph_0}\geq 2^{\aleph_0}=\aleph_5$.
On the other hand, $\aleph_5=\aleph_5^{\aleph_0}\geq \aleph_4^{\aleph_0}$.
For arbitrary cardinals $2\leq\lambda<\kappa$, $\kappa^\kappa=\lambda^\kappa$.

This is enough to find the order and specify all but one inequality, which is undecidable e.g. by Easton's theorem.