Can ZFC prove that there is a regular uncountable cardinal $\kappa$ such that $2^{<\kappa} < 2^\kappa$?
Note, if the answer is no, it would require a strong global violation of SCH, so large cardinals to force this.
On the other hand, it may be true for an easy reason, but I haven't found one. But I did figure it out under the assumption that there are no weak inaccessibles.
It was pointed out to me by Emil on mathoverflow that my argument under the assumption "no weak inaccessibles" goes through without that assumption. Here's the argument:
Assume towards a contradiction that that for all regular $\kappa$, $2^{<\kappa} = 2^\kappa$. Denote $\mathfrak{c} = 2^\omega$. We will arrive at a contradiction by showing that $2^\alpha = \mathfrak{c}$ for all $\alpha$, by induction.
Suppose this holds for all $\beta < \alpha$, and $\alpha$ is regular. By hypothesis, $2^\alpha = 2^{<\alpha} = \mathfrak{c}$. Now suppose $\alpha$ is singular. We have a general formula $2^\alpha = (2^{<\alpha})^{cf(\alpha)}$ (see Jech Chapter 5). In this case we have $2^\alpha = \mathfrak{c}^{cf(\alpha)} = (2^\beta)^{cf(\alpha)} = 2^\beta = \mathfrak{c}$ for some $\beta < \alpha$.
By the way, this generalizes to show that for every $\alpha$, there is a regular $\kappa \geq \alpha$ such that $2^\kappa > 2^{<\kappa}$.