`Show that the class of ordinals α with $$ℵ_α^{ℵ_0} = ℵ_α$$ is a proper class. Then show that the class of ordinals α with $$ℵ_α^{ℵ_0}>ℵ_α$$ is also a proper class.
What i have so far: So this one has given me lots of trouble since i cant assume the Generalised Continuum Hypothesis. But my idea is to some how use co final characters to some how show there is a surjection. But thats all i have as an idea, if anyone can offer hints that will help very much
On
Do you know the formulae for powers of cardinals?
Hausdorff formula, for all $n \in \omega$:
$$\aleph_{\alpha +n}^{\aleph_\beta} = {\aleph_\alpha}^{\aleph_\beta} \aleph_{\alpha + n}$$
Tarski formula: for $\alpha$ a limit ordinal and $\beta < cf(\alpha)$, so certainly for $\beta = 0$:
$$\aleph_\alpha^{\aleph_\beta} = \sum_{\gamma < \alpha} \aleph_\gamma^{\aleph_\beta} $$
This will maybe help evaluate the different ${\aleph_\alpha}^{\aleph_0}$, as a start: for $\alpha = 0$ we have that it equals $\aleph_0^{\aleph_0} = 2^{\aleph_0} > \aleph_0$
First note that if $\lambda=\kappa^{\aleph_0}$, then $\lambda^{\aleph_0}=\lambda$.
Secondly, recall König's theorem, and its consequence: $\kappa<\kappa^{\operatorname{cf}(\kappa)}$.