I have the following problem in front of me.
Show that if $\kappa$ is the least cardinal such that $2^\kappa>2^{\aleph_0},$ then $\kappa$ is regular.
I've scribbled this:
Suppose $$\kappa=\coprod_{\alpha<\lambda}X_\alpha,$$ where $\lambda<\kappa$ is a cardinal, and for each $\alpha<\lambda$ we have $|X_\alpha|<\kappa.$ Then $$\Large 2^{\aleph_0}<2^\kappa=2^{\coprod_{\alpha<\lambda}X_\alpha}=\prod_{\alpha<\lambda}2^{X_\alpha}=(\sup_{\alpha<\lambda}2^{X_\alpha})^\lambda.$$
If the last thing were equal to $2^\lambda$, I would get a contradiction, and I'd be done. But I have no idea what the last thing equals.
I also tried to write down some formulas with cofinality and suprema, but I didn't get any farther. What's the simplest way to do it? The thing above uses some nontrivial facts that I don't even understand very well.
Let $\lambda=\operatorname{cf}\kappa$, and let $\langle\alpha_\xi:\xi<\lambda\rangle$ be a cofinal sequence in $\kappa$. Then $|\alpha_\xi|<\kappa$ for each $\xi<\lambda$, so $2^{|\alpha_\xi|}\le 2^\omega$ for each $\xi<\lambda$. Thus, $\sup_{\xi<\lambda}2^{|\alpha_\xi|}=2^\omega$, and $(2^\omega)^\lambda=2^\lambda$ (since $\lambda\ge\omega$). And if $\lambda<\kappa$, then $2^\lambda\le 2^\omega<2^\kappa$.
Note that you don’t actually need equality in the last step of your displayed line: all you need for this result is that
$$\prod_{\alpha<\lambda}2^{|X_\alpha|}\le\left(\sup_{\alpha<\lambda}2^{|X_\alpha|}\right)^\lambda\;,$$
which is pretty easy to see.