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} $ 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?
 A: Since $2^{\aleph_4}\le\aleph_4^{\aleph_4}\le(2^{\aleph_4})^{\aleph_4}=2^{\aleph_4}$ etc. (a) and (b) are both $2^{\aleph_4}\ge\aleph_5$ (both (e) and (f), which exceed (d)). Analyze (c) using $\aleph_5=2^{\aleph_0}\le\aleph_4^{\aleph_0}\le2^{\aleph_4\aleph_0}=2^{\aleph_4}$, so dropping the brackets $a=b\ge c\ge e=f>d$. But $e=\aleph_5^{\aleph_0}\ge c$, co the second $\ge$ becomes $=$. Since $b=c$ is equivalent to $2^{\aleph_4}=2^{\aleph_0}$, it's undecidable as @spaceisdarkgreen notes (indeed, @tomasz notes Easton's theorem shows this).
A: $\aleph_4$ 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.
