Define $N_n$ as $n$ th natural number: $N_0=0, N_1=1, N_2=2, …$.
What happens after exponentiation?
We have the following equation: $2^{N_n}=N_{2^{n}}$.
(Which says: For all finite cardinal $n$ we have: $2^{n~\text{th finite cardinal}}=2^{n}~\text{th finite cardinal}$).
What this means?
The gap between $N_n$ and $2^{N_n}$ is rapidly increasing in exponential speed.
Now look at the $\text{GCH}$. It says that the gap between an infinite cardinal $\kappa$ and $2^{\kappa}$ is just a constant number $1$ in cardinals. Even in models for total failure of $\text{GCH}$ we usually have a finite gap between $\kappa$ and $2^{\kappa}$. Now if we look at the infinite cardinals as generalization of natural numbers it seems we should restate continuum hypothesis with more acceleration for the function $\kappa \mapsto 2^{\kappa}$ in order to uniform the behavior of exponentiation function in finite and infinite cardinals. Note to the following statement:
For all cardinal $\kappa$ we have $2^{\kappa~\text{th infinite cardinal}}=2^{\kappa}~\text{th infinite cardinal}$.
This is a direct generalization of the equation $\forall n\in \omega~~~~~2^{N_n}=N_{2^n}$ to the following form:
Natural Continuum Hypothesis (NCH): $~~~\forall \kappa\in Card~~~~~2^{\aleph_{\kappa}}=\aleph_{2^{\kappa}}$
Unfortunately $\text{NCH}$ is contradictory by Konig's lemma because assuming $\text{NCH}$ we have: $\aleph_{\aleph_0}<cf(2^{\aleph_{\aleph_0}})=cf(\aleph_{\aleph_1})\leq \aleph_1$ a contradiction.
(Thanks to Ramiro and Emil for their advices.)
But the acceleration problem remains open. The main question here is this:
Can we have a rapidly increasing gap between carinals by exponentiation? In the other words:
Question 1: Assuming consistency of $\text{ZFC}$ (plus some large cardinal axiom or axiom of constructibility), is the following statement consistent with $\text{ZFC}$?
$\forall \kappa\in Card~~~\exists \lambda \in Card~;~~~~~\lambda\geq 2^{\kappa}~\wedge~2^{\aleph_{\kappa}}=\aleph_{\lambda}$
Definition 1: Let $\kappa$ be a (finite or infinite) cardinal. Define $\kappa$-based beth function as follows:
$\beth_{(\kappa)}:Ord\longrightarrow Card$
$\beth_{(\kappa)}(0):=\kappa$
$\forall \alpha\in Ord~~~\beth_{(\kappa)}(\alpha +1):=2^{\beth_{(\kappa)}(\alpha)}$
$\forall \alpha\in LimitOrd~~~\beth_{(\kappa)}(\alpha):=\bigcup_{\beta \in \alpha}\beth_{(\kappa)}(\beta)$
Definition 2: Let $F:Card\longrightarrow Card$ be an increasing function and $\delta$ an ordinal. We say that $F$ has acceleration rank $\delta$ if $\delta=min\{\alpha\in Ord~|~\forall \kappa\in Card~~~\beth_{(\kappa)}(\alpha)\leq F(\kappa) < \beth_{(\kappa)}(\alpha+1)\}$.
For example the functions $\kappa\mapsto \kappa^{+}$, $\kappa\mapsto 2^{\kappa}$, $\kappa\mapsto 2^{2^{\kappa}}$ have acceleration ranks $0, 1, 2$ respectively.
Question 2: Is there any limitation for acceleration of the continuum function? Precisely is the following statement true?
"For any ordinal $\delta$ there is an increasing function $F:Card\longrightarrow Card$ with acceleration rank $\delta$ such that assuming consistency of $\text{ZFC}$ (and some large cardinal axiom) one can prove the consistency of $\text{ZFC}$ with the statement $\forall \kappa\in Card~~~2^{\aleph_{\kappa}}=\aleph_{F(\kappa)}$."
According to Emil's interesting comment I added his question here:
Question 3: Assuming consistency of $\text{ZFC}$ (and some large cardinal axiom or axiom of constructibility), is the following consistent with $\text{ZFC}$?
$\forall \kappa\in Card~~~~\exists \lambda\in Card~~~~~2^{\aleph_{\kappa}}=\aleph_{\lambda}$
Note that it is not trivial that one can have a "cardinal index" for $\aleph$ as value of $2^{\aleph_{\kappa}}$ everywhere. Perhaps we will need some non-cardinal ordinals as indexes to avoid inconsistency.
Best Answer
In the following answer, by Foreman-woodin model, I mean the model constructed by them in the paper "The generalized continuum hypothesis can fail everywhere. Ann. of Math. (2) 133 (1991), no. 1, 1–35. "
Questions 1 and 3 have positive answer: In Foreman-Woodin model for the total failure of GCH the following hold:
1) For all infinite cardinal $\kappa, 2^{\kappa}$ is weakly inaccessible, and hence a fixed point of the $\aleph-$function,
2) If $\kappa \leq \lambda< 2^{\kappa},$ then $ 2^{\lambda}= 2^{\kappa}.$
In this model for all infinite cardinals $\kappa, 2^{\kappa}=\aleph_{ 2^{\kappa}}$ in particular for all fixed points $\kappa$ of the $\aleph-$function, $2^{\aleph_\kappa}=\aleph_{ 2^{\kappa}}$. Also note that in this model for all infinite cardinals $\kappa,$ if we let $\lambda=2^{\aleph_\kappa},$ then $\lambda \geq 2^\kappa,$ and $2^{\aleph_\kappa}=\aleph_\lambda.$ So both of questions 1 and 3 have a positive answer.
For your question 2, $\delta$ can be arbitrary large: Start with GCH+there exists a supercompact cardinal $\kappa$+ there are infinitely many inaccessibles above it. Now let $\delta$ be any ordinal $<\kappa.$ Force with Foreman-Woodin construction above $\delta$ (in the sense that let the first point of the Radin club added during their forcing construction be above $\delta$). In their final model (which is $V_\kappa$ of some extension of the ground model) for all infinite cardinals $\lambda <\kappa, 2^\lambda \geq \lambda^{+\delta}$. So if $F$ is defined in the ground model by $F(\kappa)=\kappa^{+\delta},$ then the acceleration rank of $F$ is $\delta$ (using GCH), and in the finial model for all infinite cardinals $\kappa, 2^\kappa \geq F(\kappa).$
Remark. I may note that we can not define the function $F$ in the ground model, such that it is the realization of power function in the extension, but we can find some inner model of the final extension in which $GCH$ holds and such a function $F$ is definable.