Let $\kappa < \beth_2$ and $\lambda<\beth_1$ be cardinals. What can we say about $\kappa^{\lambda}$ without assuming CH? Is it true that $\kappa^{\lambda} < \beth_2$ or $\kappa^{<\beth_1} < \beth_2$?
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2$\begingroup$ The title was changed to something I disagree with, because "exponent" refers to a quantity denoted by a superscript, and we are not concerned with inequalities between these, but rather with values of exponentiation as a binary operation on cardinals. Thus, I think the original title was more correct than the so-called "improvement" by user64494 -- even if the plural of "exponentiation" sounds awkward to some ears. $\endgroup$– Todd TrimbleCommented Jun 7, 2019 at 18:39
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$\begingroup$ @ToddTrimble I agree and have rolled back. $\endgroup$– Noah SchweberCommented Jun 7, 2019 at 20:44
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Even if CH holds, this can break.
Suppose $\beth_1=\aleph_1$ and $\beth_2=\aleph_{\omega+1}$. Let $\lambda=\aleph_0$ and $\kappa=\aleph_\omega$. By Konig's theorem we know that $\kappa^\lambda=\kappa^{cf(\kappa)}>\kappa$; since $\kappa^+=\beth_2$ this means $\kappa^\lambda\ge\beth_2$ (which in turn means $\kappa^\lambda=\beth_2$, of course).
answered Jun 7, 2019 at 15:22