For any cardinal numbers $a\neq b$, is there a cardinal number $c$ such that $c^a\neq c^b$?

Question

For any cardinal numbers $a\neq b$, is there a cardinal number $c$ such that $c^a\neq c^b$?

If there is no (or ZFC undecidable) such cardinal number, then is the following true? Let $a,b$ be two cardinal numbers, if $c^a=c^b$ for any cardinal number $c$, then $a=b.$

This is motivated by the question here when I consider the category of set.

Answer 1

Yes. Recall the $\beth$ numbers:

1. $\beth_0=\aleph_0$;
2. $\beth_{\alpha+1}=2^{\beth_\alpha}$; and
3. $\beth_\delta=\sup\{\beth_\gamma\mid\gamma<\delta\}$ for a limit ordinal $\delta$.

So if $\delta$ is a limit ordinal, $\beth_\delta$ is a strong limit cardinal, and moreover, if $\lambda$ is the cofinality of $\beth_\delta$ (or that of $\delta$), then $\beth_\delta^\gamma=\beth_\delta$ for all $\gamma<\lambda$. But, recall König's theorem and its consequence: if $\gamma\geq\lambda$, then $\beth_\delta^\gamma>\beth_\delta$.

Now. If $\kappa<\lambda$, then there is a regular cardinal $\mu$ such that $\kappa<\mu\leq\lambda$, e.g. $\kappa^+$. Therefore $\beth_\mu$ is a cardinal such that:

1. $\beth_\mu^\kappa=\beth_\mu$, and
2. $\beth_\mu^\lambda\geq\beth_\mu^\mu>\beth_\mu$.