Question about $\kappa^{<\kappa} = \kappa$ Set theory noob here...
So I found a few results stating that if a cardinal $\kappa$ fulfills some property X, then $\kappa^{<\kappa} = \kappa$
(e.g. X is '$\kappa$ is strongly inaccessible', see If $ \kappa $ is strongly inaccessible, then $ \kappa^{<\kappa} = \kappa $ ).
My question concerns the other implication direction. If we assume that $\kappa^{<\kappa} = \kappa$ for some $\kappa$, what can we say about the "size" of $\kappa$? Is there an informal way to think about such cardinals? (maybe assume that $\kappa$ is uncountable)
 A: The following are equivalent for an infinite cardinal $\kappa$:

*

*$\kappa^{<\kappa} = \kappa$

*$\kappa$ is regular and for all $\lambda < \kappa$, $2^\lambda \leq \kappa$.

*(a) $\kappa = \aleph_0$ or (b) $\kappa$ is (weakly) inaccessible with $2^\lambda\leq \kappa$ for all $\lambda < \kappa$, or (c) $\kappa = \mu^+$ and GCH holds at $\mu$: $2^\mu = \kappa$.

Proof: $1\implies 2$: Assume $\kappa^{<\kappa} = \kappa$. Suppose for contradiction that $\kappa$ is singular. Then $\kappa^{<\kappa} \geq \kappa^{\mathrm{cf}(\kappa)} > \kappa$. Now suppose for contradiction that there exists $\lambda<\kappa$ such that $2^\lambda > \kappa$. Then $\kappa^{<\kappa}\geq \kappa^\lambda \geq 2^\lambda > \kappa$.
$2\implies 3$: Assume $\kappa$ is regular and for all $\lambda < \kappa$, $2^\lambda \leq \kappa$. If $\kappa$ is a limit, then it is a regular limit, so it is $\aleph_0$ or inaccessible. If $\kappa = \mu^+$ is a successor, then $2^\mu \leq \kappa$, but also $2^\mu>\mu$ implies $2^\mu\geq \kappa$. So $2^\mu = \kappa$.
$3\implies 1$: (a) $\aleph_0^n = \aleph_0$ for all finite $n$, so $\aleph_0^{<\aleph_0} = \aleph_0$.
(b) Suppose $\kappa$ is an uncountable regular limit cardinal with $2^\lambda\leq \kappa$ for all $\lambda < \kappa$. Fix $\mu$ with $\aleph_0\leq \mu<\kappa$. Since $\kappa$ is regular, every function $\mu\to \kappa$ has bounded range. So $\kappa^\mu = \sup_{\mu\leq \lambda < \kappa} \lambda^\mu$. Now $\lambda^\mu \leq (2^\lambda)^\lambda = 2^\lambda \leq \kappa$, so $\kappa^\mu \leq \sup_{\mu\leq \lambda < \kappa} \kappa = \kappa$. Thus $\kappa^{<\kappa} = \kappa$.
(c) If $\kappa = \mu^+$ and $2^\mu = \kappa$, then $\kappa^{<\kappa} = \kappa^\mu = (2^\mu)^\mu = 2^\mu = \kappa$.

So what the condition $\kappa^{<\kappa} = \kappa$ tells us about the "size" of $\kappa$ is entirely dependent on the whims of the continuum function.
At one extreme, if GCH holds, then every regular cardinal satisfies $\kappa^{<\kappa} = \kappa$.
At the other extreme, it is possible that there are no uncountable cardinals satisfying $\kappa^{<\kappa} = \kappa$. It is a theorem of Woodin (see Foreman and Woodin The GCH can fail everywhere) that assuming a certain large cardinal hypothesis is consistent, we can find a model of set theory in which $2^\kappa = \kappa^{++}$ for all infinite cardinals $\kappa$. In this model, every limit cardinal is a strong limit, so every inaccessible cardinal is strongly inaccessible. Then cutting off the model at $V_\kappa$, where $\kappa$ is the least inaccessible cardinal in the model, gives a new model in which GCH fails everywhere and there are no inaccessible cardinals, so there are no cardinals satisfying (b) or (c) in the characterization  above.
