Is $2^ \kappa \leq \kappa ^ {\operatorname{cf}(\kappa)}$ for infinite cardinals $\kappa$? Definitions
The cofinality of an ordinal is defined to be the least ordertype of cofinal subsets.
A cardinal is defined to be an ordinal which is not equinumerous to any smaller ordinal.
Work in ZFC.
The Question

Is $2^ \kappa \leq \kappa ^ {\operatorname{cf}(\kappa)}$ for infinite cardinals $\kappa$? If not, what extra assumptions would we have to add to be able to say this?

I know that if the Generalized Continuum Hypothesis (GCH) holds then we actually have $2^ \kappa = \kappa^{\operatorname{cf}(\kappa)}$, which in my opinion has a certain elegance to it, so I was wondering how much we can say in more general circumstances.
Obviously if $\kappa= \operatorname{cf}(\kappa)$ is an infinite regular cardinal we have $2^ \kappa=\kappa^ \kappa= \kappa^{\operatorname{cf}(\kappa)}$, but what about when $\kappa$ is singular?
 A: Not necessarily. We always have $\kappa^{\operatorname{cf}\kappa}\le 2^\kappa,$ and, as you note, equality holds when $\kappa$ is regular (and it holds always for GCH). However, it is consistent that it fails for singular cardinals.
Start with a model of GCH, and use Cohen forcing to add $\aleph_{\omega+2}$ subsets of $\aleph_1$ so that $2^{\aleph_1}=\aleph_{\omega+2}$ in the forcing extension. (In other words, the forcing poset is the set of countable partial functions from $\aleph_{\omega+2}\times \aleph_1\to 2.$) Then, we have $2^{\aleph_\omega}\ge 2^{\aleph_1}=\aleph_{\omega+2}$ (actually, equality holds here, but that's more than we need to know). However, the forcing is countably closed, so it doesn't add any new countable subsets of $\aleph_\omega,$ so we still have $\aleph_\omega^{\aleph_0}=\aleph_{\omega+1}.$
More generally, while it's easy to mangle the continuum function any which way, it is hard (i.e. requires some large cardinals and more elaborate forcing constructions) to violate the SCH, which says that if $2^{\operatorname{cf}\kappa}<\kappa,$ then $\kappa^{\operatorname{cf}\kappa}=\kappa^+$.
