Let $\varepsilon$ be an epsilon number, i.e., an ordinal such that $\omega^{\varepsilon} = \varepsilon$. Is it true that then, for each pair $\alpha,\beta$ of ordinals, with $\alpha,\beta < \varepsilon$, we have $\alpha^\beta < \varepsilon$?
Does the converse also hold? If $\varepsilon$ is an infinite ordinal unreachable from below with exponentiation, is it an epsilon number?
I suspect this is the case (I haven't seen it written anywhere) but I'm having trouble to prove it. Ordinal exponentiation doesn't behave so nicely.
If there exist $\alpha, \beta < \varepsilon$ with $\alpha^\beta = \varepsilon$, we can show that the Cantor Normal Form of $\varepsilon$ has at least two terms, but I don't know what to do if we have $\alpha^\beta \geq \varepsilon$...