A choice-free cardinal equality

For a cardinal $\mathfrak{a}$, let $\mathrm{f}\mathrm{i}\mathrm{n}(\mathfrak{a})$ be the cardinality of the set of all finite subsets of a set which is of cardinality $\mathfrak{a}$. It is proved without the aid of the axiom of choice that for all infinite cardinals $\mathfrak{a}$ and all natural numbers $n$, $$2^{\mathrm{f}\mathrm{i}\mathrm{n}(\mathfrak{a}{)}^n}=2^{[\mathrm{f}\mathrm{i}\mathrm{n}(\mathfrak{a}){]}^n}\mathrm{.}$$ On the other hand, it is proved that the following statement is consistent with $\mathsf{Z}\mathsf{F}$: there exists an infinite cardinal $\mathfrak{a}$ such that $$2^{\mathrm{f}\mathrm{i}\mathrm{n}(\mathfrak{a})}<2^{\mathrm{f}\mathrm{i}\mathrm{n}(\mathfrak{a}{)}^2}<2^{\mathrm{f}\mathrm{i}\mathrm{n}(\mathfrak{a}{)}^3}<\cdots <2^{\mathrm{f}\mathrm{i}\mathrm{n}(\mathrm{f}\mathrm{i}\mathrm{n}(\mathfrak{a}))}\mathrm{.}$$