I'm wondering if $\lambda = (2^{\mathfrak{c}})^{<2^\mathfrak{c}} = \sup_{\kappa < 2^\mathfrak{c}} 2^{\mathfrak{c}\cdot \kappa}$ can be calcuted without invoking the general continuum hypothesis and related. So far I can see $\lambda \geq 2^\mathfrak{c}$ and equality holds under GCH.
More precisely I want to know if ZFC proves $\lambda = 2^\mathfrak{c}$.
This cardinal appeared when I tried to calculate the amount of pseudocompact subspaces of $\beta\mathbb{N}$.
It could be anything. Suppose that $\mathfrak c=\aleph_1$, $2^{\aleph_1}=\aleph_3$, and $2^{\aleph_2}=\mu$, for any $\mu$, such that $\operatorname{cf}(\lambda)\geq\aleph_3$.
Then $\lambda=\mu$ in that case. Since $\mu$ can be as small or as large as you'd like it to be, with the obvious constraint on cofinality, there's no good way of saying what it may or may not be.
Two caveats here are:
If $2^\frak c$ is a successor cardinal, say $\mu^+$, then $\lambda=2^\mu$.
If $2^\frak c$ is a limit cardinal, then it must have uncountable cofinality, and so the question is really about the behaviour of the continuum function below $2^\frak c$.