TITLE: weak consequence of GCH
QUESTION [7 upvotes]: 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.

REPLY [5 votes]: 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 $2^\kappa > 2^{<\kappa}$.