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 the Dedekind finite sets are not necessarily closed under exponentiation.