For which cardinal numbers $\kappa$ is it consistent with ZFC that $\kappa^{\mathrm{cf}(\kappa)} < \kappa^\kappa$? 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).

Question. For which cardinal numbers $\kappa$ is it consistent with ZFC that $\kappa^{\mathrm{cf}(\kappa)} < \kappa^\kappa$?

Clearly no such $\kappa$ can be regular, and $\kappa = \beth_\omega$ will not work, but that's all I know.
 A: Start with a model of GCH.  Let $\kappa$ be singular.  Add $\kappa^{++}$ Cohen subsets of $\mathrm{cf}(\kappa)^+$-- the forcing is $\mathrm{cf}(\kappa)^+$-closed and $\mathrm{cf}(\kappa)^{++}$-c.c.  Then we'll have $\kappa^\kappa = \kappa^{++}$ but $\kappa^{\mathrm{cf}(\kappa)} = \kappa^+$ since we've added no $\mathrm{cf}(\kappa)$-sequences of ordinals.
A: (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 $\kappa=\aleph_\omega$.  
