The following question was posed by Guozhen Shen comment on a prior posting of mine.
Does the statement that there is at most one cardinal between an infinite cardinal and its powerset implies the axiom of choice in $\sf ZF$?
However before reaching to this one can solve simpler versions before it, for instance a related statement similar to the $\sf GCH$ in the following strong manner:
$\forall \text { infinite cardinals } \kappa, \lambda \, ( \kappa < \lambda \to 2^\kappa \leq \lambda)$
Its important to notice that those cardinals are Scott's cardinals, so they are not necessarily well orderable (meaning their elements are well-orderable).
The proof that this implies $\sf AC$ over $\sf ZF$ is actually very easy. Simply take $\kappa + \aleph(\kappa)$, if $\kappa$ is not well-orderable, then $\kappa < \kappa+\aleph(\kappa)$ [By definition of Hartogs] , so $2^\kappa \leq \kappa + \aleph(\kappa)$ but since $\kappa < 2^\kappa$, then this entails $2^\kappa \leq \aleph(\kappa)$ (Specker: Lemma 2.3), thus $\kappa$ is well orderable. A contradiction, so $\kappa$ must be well-orderable.
We notice that this proof is much simpler than the weaker form of $\sf GCH$ state by absence of intervening cardinals between cardinalities of infinite sets and their powersets.
Similarly we can have a strong form of Guozhen Shen question, that of:
$\forall \text { infinite cardinals } \kappa, \lambda, \zeta: \kappa \leq \lambda < \zeta \to 2^\kappa \leq \zeta$
Also the proof of $\sf AC$ from this in $\sf ZF$ is very simple, we start wit the same proof of the strong $\sf GCH$, but add here the cardinal $\kappa + (\aleph(\kappa))^+$ or what I'd label as $\kappa + \aleph_1(\kappa)$, clearly we have $\kappa < \kappa + \aleph(\kappa) < \kappa + \aleph_1(\kappa)$, and so we must have $2^\kappa \leq \kappa + \aleph_1(\kappa)$, and $2^\kappa \leq \aleph_1(\kappa)$ follows by Lemma 2.3.
However, the weaker statements of $\sf GCH$ and the maximally single interval powerset phrased in terms of intervening cardinals between sets and their powers (i.e.; Shen's question), are harder to prove. The proof of the first goes like that: Lets take $X$ to be an infinite stage $V_\alpha$ of the cumulative hierarchy of $\sf ZF$. Let it be some limit stage. So, here we'll have $X \times X \subset X$, and so $|X \times X|= |X|$, let $\aleph(X)$ be defined in the original Hartog manner as the set of all equivalence classes under order isomorphism of well-orderings of subsets of $X$, this is a subset of $\mathcal P^2(X)$, so we have $\aleph(X) \leq |\mathcal P^2(X)|$.
Now, we have $|\mathcal P(X)| + \aleph(X) \geq |\mathcal P(X)| $ (Definition of $\geq$),
- if $|\mathcal P(X)| + \aleph(X) > |\mathcal P(X)| $ then
$|\mathcal P(X)| + \aleph(X)$ would intervene between $|\mathcal P(X)|$ and $|\mathcal P^2(X)|$ (Specker: Lemma 2.3);
- if $|\mathcal P(X)| + \aleph(X) = |\mathcal P(X)| $ then $\aleph(X) < |\mathcal P(X)|$ (otherwise $\mathcal P(X)$ is well-orderable). But, by then we'll have $ |X|+\aleph(X) $ intervening between $|X|$ and $|\mathcal P(X)|$(Definition of Hartogs; Specker: Lemma 2.3).
The above argument would constitute the nucleus of further proofs here, I describe this by $\aleph(X)$ causing a waver across $\mathcal P(X)$! That is, it induces an intervening cardinal across $|\mathcal P(X)|$, i.e. between cardinalities of $\mathcal P(X)$ and either of $X$ or $\mathcal P^2(X)$. For a hartog $\aleph(X)$ we name the least $\mathcal P^n(X)$ that its injective to as the ceiling, while its floor denotes $X$ itself where it is not injective to. Potentially a Hartog would induce a wavering state across all single power intervals below its ceiling and above its floor.
I'm not sure using this wavering scenario can solve the question I posed, but it might be of help?
Now, before solving Shen's question we can solve a simpler one, that if $\sf ZF$ proves $\sf AC$ to follow from the statement that no more than two intervening cardinals between $X$ and $\mathcal P^2(X)$ can exist if $X$ is infinite.
We repeat the same scenario done above but also bring $\aleph(\mathcal P(X)); \aleph(\mathcal P^2(X))$ into the picture, these would cause intervening cardinals that waver across $\mathcal P^2(X)$ and across $\mathcal P^3(X)$, the result of these wavers is that we'll have three cardinalities intervening between some set and its second powerset.
To be noted is that we used the theorem of $\sf ZF$ stating that if $|X \times X|=|X|$ then $|\mathcal P(X) \times \mathcal P(X)|=|\mathcal P(X)|$. This way we prove by induction that actually all stages $V_\alpha$ of the cumulative hierarchy to be bijective to their Cartesian products.
The hard Guozhen Shen question, I think can be solved by bringing more Hartogs into the picture. I'll use the notation $\aleph_{n+1}(X)$ to denote $(\aleph_n(X))^+$. So, we bring $\aleph_1(X)$, and $\aleph_1(\mathcal P(X))$, now the first is injective to $\mathcal P^3(X)$, and the second to $\mathcal P^4(X)$ (because both $X$ and $\mathcal P(X)$ are bijective to their Cartesian products: see Glazer's comment). These would cause wavers across $\mathcal P^3(X); \mathcal P^2(X)$. So, we have five Hartogs $\{\aleph(X), \aleph(\mathcal P(X)), \aleph(\mathcal P^2(X)), \aleph_1(X), \aleph_1(\mathcal P(X))\}$ causing wavers across three levels $\{\mathcal P(X), \mathcal P^2(X), \mathcal P^3(X)\}$ and four single power intervals, so one interval must have two intervening cardinals in some single power interval.