ZFC proves that $\kappa^{\mathrm{cf}(\kappa)} \leq \kappa^\kappa$ for all infinite cardinal numbers $\kappa$. Further, it is consistent with ZFC that we always have equality (e.g. assume GCH).



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> **Question.** For which cardinal numbers $\kappa$ is it consistent with ZFC that $\kappa^{\mathrm{cf}(\kappa)} < \kappa^\kappa$?
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Clearly no such $\kappa$ can be regular, and $\kappa = \beth\_\omega$ will not work, but that's all I know.



For which cardinal numbers $\kappa$ is it consistent with ZFC that $\kappa^{\mathrm{cf}(\kappa)} < \kappa^\kappa$?
(A very partial answer only, but too long for a comment.) 


In light of your second comment under Monroe's answer, I interpret your question as follows: 


Given a formula $\varphi(x)$, write $\kappa\_\varphi$ for the least cardinal satisfying $\varphi$. ($\kappa\_\varphi$ could be undefined, of course. But you are only concerned with formulas which are satisfied by a unique cardinal.)


Write $C(\kappa)$ to abbreviate $\kappa^{cf(\kappa)} < \kappa^{\kappa}$. 


How can we find out if "ZFC   +   $\kappa\_\varphi$ exists   +   $C(\kappa\_\varphi)$" is consistent? 


A sufficient condition is the following: $\varphi$ is absolute under cardinal-preserving extensions (so in particular, $\varphi$ may use ordinal arithmetic, cardinal successor, the $\aleph$-function, etc), ZFC proves that $\kappa\_\varphi$ exists and is singular. (This is really just a reformulation of Monroe Eskew's answer.)


An example is the formula $\kappa=\aleph\_\omega$. 


