A New Continuum Hypothesis (Revised Version) 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.
 A: NCH is inconsistent with ZFC:
First, under NCH, we have:
$$
2^{\aleph_{\aleph_{\aleph_0}}} =
\aleph_{2^{\aleph_{\aleph_0}}} =
\aleph_{\aleph_{2^{\aleph_0}}} =
\aleph_{\aleph_{\aleph_1}} .
$$
Under NCH, $\aleph_{\aleph_{\aleph_0}}$ is strong limit, since, for any $n \in \omega$:
$$
2^{\aleph_{\aleph_n}} =
\aleph_{2^{\aleph_n}} =
\aleph_{\aleph_{2^n}} <
\aleph_{\aleph_{\aleph_0}} .
$$
Now, consider the Theorem 7.3 in Chapter 14 (Cardinal Arithmetic - Abraham and Magidor) of the Handbook of Set Theory (page 1210):

Suppose that $\delta$ is a limit ordinal and
  ${|\delta|}^{\operatorname{cf}(\delta)} < \aleph_{\delta}$. Then
  $$
{\aleph_{\delta}}^{\operatorname{cf}(\delta)} <
\aleph_{({|\delta|}^{+4})} .
$$

Then we have, under NCH:
$$
2^{\aleph_{\aleph_{\aleph_0}}} =
{(\aleph_{\aleph_{\aleph_0}})}^{\aleph_0} <
\aleph_{{(\aleph_{\aleph_0})}^{+4}} =
\aleph_{\aleph_{\omega + 4}} .
$$
However,
$$
\aleph_{\aleph_{\aleph_1}} >
\aleph_{\aleph_{\omega + 4}} .
$$
A: As for question 2, if $\forall\kappa\,2^{\aleph_\kappa}=\aleph_{F(\kappa)}$, then $F$ does not have an acceleration rank.
If the rank were $\delta\ge2$, we could take $\kappa$ to be any fixpoint of the $\aleph$ function to obtain a contradiction:
$$2^{\aleph_\kappa}=2^\kappa\le\aleph_{2^\kappa}=\aleph_{\beth_{(\kappa)}(1)}.$$
On the other hand, a finite acceleration rank is impossible: this would imply $F(n)<\omega$ for every $n<\omega$, i.e., $\aleph_\omega$ is strongly limit. In particular, $\omega\le F(\omega)\le\beth_{(\omega)}(\delta+1)=\beth_{\delta+1}<\aleph_\omega$, hence $\mathrm{cf}(2^{\aleph_\omega})=\mathrm{cf}(\aleph_{F(\omega)})=F(\omega)<\aleph_\omega$, contradicting König’s lemma.
A: Let me provide a large cardinal lower bound for the statement in question 1. 
Since $\kappa\lt\kappa+1\lt 2^\kappa$ for $\kappa>1$, the statement implies that there is
a universal failure of the GCH above $\aleph_1$ at cardinals of the form $\aleph_\kappa$ for $\kappa$ a cardinal. In particular, this implies that
the negation of the singular cardinals hypothesis
SCH. Since $\neg\text{SCH}$ is known to have large cardinal strength, we have large cardinal
lower bounds for the consistency strength of your statement.
Meanwhile, since the statement also implies violations of the SCH arbitrarily
high in the cardinals, it follows by a result of Solovay that it is inconsistent with the existence of strongly compact cardinals.
A: 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.
