It is known that the sum and the product of two Dedekindfinite cardinals are also Dedekindfinite cardinals. What about cardinal exponentiation ? Question: Let A and B be two Dedekindfinite cardinals, let C be the cardinal A power B (i.e:let x be a set with cardinal A and y be a set with cardinal B and let C be the cardinal of the set of functions with domain y and range a subset of x). Is it true that C is a Dedekindfinite cardinal ? Gérard Lang
The answer is no, not necessarily, because if there are infinite Dedekind finite sets, then the class of Dedekind finite sets is not closed under power set, and hence not closed under $A\mapsto 2^A$. To see this, simply note that if $A$ is any infinite set, then $P(A)$ has the singletons, the doubletons, the subsets of size $3$, and so on. So we can find a countably infinite subset of $P(P(A))$, and so $P(P(A))$ is not Dedekind finite. In particular, if $A$ is Dedekind finite but infinite, then $2^{2^A}$ is not Dedekind finite, and so it is consistent with ZF that the Dedekind finite sets are not closed under exponentiation. 


It is a theorem of Kuratowski that the following holds:
As Joel says, the power set of an infinite set can always be mapped onto $\omega$, therefore the secondpower set is always Dedekindinfinite. So if the first one is not, then the second one is. It should be remarked that both cases can be realized in terms of consistency.
So we can say that Dedekindfinite cardinals are closed under exponentiation if and only if every Dedekindfinite set is finite. But given two infinite Dedekindfinite sets, we cannot necessarily say that their exponentiation is Dedekindinfinite. 

