Assuming the axiom of choice the following argument is simple, for infinite $A$ it holds: $$2\lt A\leq2^A\implies 2^A\leq A^A\leq 2^{A\times A}=2^A.$$


However without the axiom of choice this doesn't have to be true anymore. For example if $A$ is an amorphous set (infinite set that cannot be written as a disjoint union of two infinite sets), then it is actually true that $2^A<3^A<4^A<\ldots< A^A$. The reason these inequalities hold is that $A^A$ is actually Dedekind-finite, so whenever we remove elements we strictly decrease in cardinality.


Of course there are still sets that obey the equation $A^A=2^A$, even if $A$ cannot be well-ordered. For example given any set $A$ it is not hard to verify that $A^\omega$ has the property $A^\omega\times A^\omega=A^\omega$. From this follows:


$$2\lt A^\omega\leq 2^{A^\omega}\implies 2^{A^\omega}\leq\left(A^\omega\right)^{A^\omega}\leq \left(2^{A^\omega}\right)^{A^\omega}=2^{A^\omega}$$
(In fact we can replace $\omega$ by any set $\tau$ such that $\tau+\tau=\tau$)


But I have a hard time to believe that these two things are equivalent: $$A^A=2^A\iff A\times A=A.$$



> 
> **Question I.** Is there anything known on the properties of sets for which $A^A=2^A$?
> 
> 
> **Question II.** If $2^A=A^A$ does not characterize the sets for which $A\times A=A$, does the axiom "*For every infinite $A$, $2^A=A^A$*", imply the axiom of choice?
> 
> 
> 


If the answer is unknown, does this question (or variants, or closely related questions) appeared in the literature?


It seems like a plausible question by Tarski or Sierpiński. I found several other questions I have asked before to be questions that have been asked in one a paper or another.



When does $A^A=2^A$ without the axiom of choice?
David Pincus in "[A note on the cardinal factorial](https://eudml.org/doc/215090)" (Fundamenta Mathematicae vol.98(1), pages 21-24(1978)) proves that $A^A=2^A$ does not imply the axiom of choice, therefore it does not characterise the sets for which $A=A\times A$. The counterexample is the model from his paper "[Cardinal representatives](https://doi.org/10.1007/BF02760841)", Israel Journal of Mathematics, vol.18, pages 321-344 (1974). 


As Pincus writes on the last lines of [Pincus78]:



> 
> c. Our arguments [ in the proof of "4.The Main Theorem; If $x$ is infinite then $2^x=x!$" ] have made little use of the particular definition of $x!$. Indeed let $\mathcal{F}$ be any set valued operation which satisfies: 
> 
> 
> (1) The predicate $y\in\mathcal{F}$ is absolute from $M$ to $V$.
> 
> 
> (2) $\mathsf{ZF}$ proves $|y|\leq x \Rightarrow |\mathcal{F}(y)|\leq |\mathcal{F}(y)|$ and $|2x|=|x|\Rightarrow 2^x\leq|\mathcal{F}(x)|$ for infinite $x$.
> 
> 
> (3) $\mathsf{ZFC}$ proves $2^x=|\mathcal{F}(x)|$ for infinite $x$. 
> 
> 
> The statement "For every infinite $x$, $2^x=|\mathcal{F}(x)|.$", holds
>  in $M$ (and therefore is not an equivalent to the axiom of choice).
>  Examples of $\mathcal{F}$, apart from $x!$, are $x^x$ and $x^x-x!$.
> 
> 
> 


Therefore, in this Pincus model, $\mathsf{ZF}$ + $\lnot\mathsf{AC}$ + "for all infinite x, $2^x=x!=x^x=x^x-x!=|\mathcal{F}(x)|$" holds for any set valued operator $\mathcal{F}$ as above.


I should say that, after failing to come up with an answer myself, I found out about this by searching into my good old friend the "Consequences of the axiom of choice" by Howard and Rubin, [Form 200](https://books.google.com/books?id=ffXxBwAAQBAJ&pg=PA50), and [Note 64](https://books.google.com/books?id=ffXxBwAAQBAJ&pg=PA265). I try to reference this book whenever I can :)


