cardinals such that $\alpha < \alpha ^ \beta$ Let $F$ be a (class) function that assigns at each cardinal $\alpha$ the minimum cardinal $\beta$ such that $\alpha < \alpha ^ \beta$. I have to prove that $\beta \in \text{Im}(F)$ iff $\beta$ is regular (i.e. $\text{cof}(\beta) = \beta)$.
I proved the "only if" part, in fact if $\beta = F(\alpha)$ is not regular then let $\mu < \beta$ be it's cofinality. This means we can write $\beta = \sum_{i \in \mu} \beta_i$, with $\beta_i <\beta$ for each $i \in \mu$. So we have: $$\alpha <\alpha^\beta = \alpha ^ {\sum_{i \in \mu} \beta_i} = \prod_{i \in \mu} \alpha^{\beta_i} = \prod_{i \in \mu} \alpha = \alpha^\mu$$
Where the third equivalence is due to the fact that $\beta$ is the minimum s.t. $\alpha < \alpha^\beta$, this is absurd because $\mu<\beta$ but $\alpha < \alpha^\mu$.
What about the "if" part?
 A: Recall that the beth ($\beth$) numbers are defined by transfinite recursion as $\beth_0=0$, $\beth_{\alpha+1}=2^{\beth_\alpha}$ and $\beth_\lambda=\sup_{\alpha<\lambda}\beth_\alpha$ for limit ordinals $\lambda$. Note that the sequence of $\beth$ numbers is strictly increasing.
It suffices to show that if $\kappa$ is regular, then $F(\beth_\kappa)=\kappa$.
To see this, note first that $\beth_\kappa$ has cofinality $\kappa$ and therefore (by König's theorem) $(\beth_\kappa)^\kappa>\beth_\kappa$.
Now, suppose $\lambda<\kappa$. We want to show that $(\beth_\kappa)^\lambda=\beth_\kappa$.
Note that if $f\!:\lambda\to\beth_\kappa$ then, since $\lambda<\kappa$, $f$ is bounded and therefore $$ {}^\lambda\beth_\kappa=\bigcup_{\alpha<\kappa}{}^\lambda\beth_\alpha=\bigcup_{\lambda<\alpha<\kappa}{}^\lambda\beth_\alpha,$$ where ${}^AB$ is the set of functions from the set $A$ to the set $B$. Thus, since $\lambda<\alpha\le\beth_\alpha$, $$(\beth_\kappa)^\lambda \le \sum_{\lambda<\alpha<\kappa}(\beth_\alpha)^\lambda \le\sum_{\lambda<\alpha<\kappa}(\beth_\alpha)^{\beth_\alpha}= \sum_{\lambda<\alpha<\kappa}\beth_{\alpha+1}=\beth_\kappa.$$
