Cardinal exponentiation without generalized continuum hypothesis

First I have to confess that I don’t know about set theory language.

Let $ A$ and $ B$ be infinite cardinals with $ A>B$ .

My question is: $ A^B=A$ ? (without assuming generalized continuum hypothesis)

Remark: assuming generalized continuum hypothesis (GCH briefly), this can be proved by the following (at least for unlimit cardinal).

Sps $ A$ is a unlimit cardinal. Then $ A=2^C$ for some $ C\ge B$ by GCH. Therefore $ A^B = (2^C)^B=2^{CB}=2^C=A$ .

Unfortunately, I don’t know how to prove for limit cardinal case. Please somebody help me!

Does measurability of cardinal $\kappa$ imply measurability of $2^\kappa$?

A cardinal $ \kappa$ is real-valued measurable if there is a $ \kappa$ -additive probability measure on $ 2^\kappa$ which vanishes on singletons. The existence of measurable $ \kappa$ is independent of ZFC.

Question: if $ \kappa$ is assumed to be real-valued measurable, does it necessarily follow that $ 2^\kappa$ is real-valued measurable?

Regarding the definition of a Reinhardt cardinal being inconsistent with choice

Consider the following comment Noah Schweber made to Embiggening regarding his (Embiggening’s) mathstackexchange question, “Could Reinhardt cardinals exist even if all cardinals below it were well-orderable”:

When we say that a Reinhardt cardinal is inconsistent with choice, we mean that if $ M$ is a model of $ ZF$ , $ \kappa$ $ \in$ $ M$ , and $ M$ $ \vDash$ $ \kappa$ is a Reinhardt cardinal,” then there is some set in $ M$ which is not well-orderable in $ M$ . But this set won’t itself be $ \kappa$ .”

It would seem reasonable to assume that this definition of a Reinhardt cardinal being inconsistent with choice would also hold for $ NGB$ .

My question is simply this: if such a model of $ NGB$ were to exist, in what way would Choice have to fail in order for $ \kappa$ in “$ \kappa$ is a Reinhardt cardinal” to be well-orderable? What sort of models of $ NGB$ would satisfy these constraints?

The Parovichenko cardinal, is it equal to $\max\{\aleph_2,\mathfrak p\}$?

Let us define the Parovichenko cardinal $ \mathfrak{P}$ as the smallest cardinal $ \kappa$ such that each compact Hausdorff space $ K$ of weight $ w(K)<\kappa$ is the continuous image of the remainder $ \beta\mathbb N\setminus\mathbb N$ of the Stone-Cech compactification of the discrete space of positive integers $ \mathbb N$ .

By a classical theorem of Parovichenko, $ \mathfrak P\ge\aleph_2$ .

On the other hand, Theorem 2.7 of this paper of van Douwen and Przymusinski implies that $ \mathfrak P\ge\mathfrak p$ where $ \mathfrak p$ is the well-known pseudointersection number.

These two results imply that $ \mathfrak P\ge\max\{\aleph_2,\mathfrak p\}$ .

So, under CH we have $ \mathfrak P=\aleph_2>\mathfrak c=\mathfrak p=\aleph_1$ .

Alan Dow and KP Hart observed (on the page 1833 of their paper in TAMS) that in the Cohen model $ \mathfrak P=\aleph_2=\mathfrak c>\mathfrak p=\aleph_1$ .

Finally, PFA implies $ \mathfrak P=\aleph_2=\mathfrak c=\mathfrak p>\aleph_1$ , see Corollary 4.6 in Baumgartner’s survey “Applications of the Proper Forcing Axiom” in the “Handbook of Set-Theoretic Topology”.

Problem. Is it consistent that $ \mathfrak P>\max\{\aleph_2,\mathfrak p\}$ ?