TITLE: A cardinal $\kappa$ such that $2^{\lambda}<\kappa$ for all $\lambda<\kappa$ is regular?
QUESTION [5 upvotes]: A cardinal $\kappa$ such that $2^{\lambda}<\kappa$ for all $\lambda<\kappa$ is regular?
I would appreciate very much an answer

REPLY [8 votes]: No. This is just the definition of a strong limit cardinal. It does not have to be regular.
For example, $\beth_\omega$. The $\beth$ numbers are defined as:
$\beth_0=\aleph_0$; $\beth_{\alpha+1}=2^{\beth_\alpha}$; and for a limit $\beta$, $\beth_\beta=\sup\{\beth_\alpha\mid\alpha<\beta\}$. It is not difficult to see that for any limit ordinal $\delta$, $\beth_\delta$ is a strong limit cardinal.