A cardinal inequality for finiteness

Question

Nearly ten years ago, I explained in a blog post that, assuming only ZF, a cardinal number $\mathfrak{n}$ is finite if and only if it satisfies this monstrous inequality: $$2^{2^{2^{2^{\mathfrak{n}}}}} \lt \left(2^{2^{2^{2^{\mathfrak{n}}}}}\right)^2 = 4^{2^{2^{2^{\mathfrak{n}}}}}$$ Where "finite" is meant in the strictest Tarskian sense: in bijection with a finite ordinal.

My question is a bit vague since "simple" has a lot of interpretations but here it is:

Is there a simpler cardinal inequality that is equivalent to finiteness and uses only cardinal exponentiation?

Answer 1

Läuchli proved in 1961 that $\mathfrak{n}$ is finite if and only if $2^{2^{\mathfrak{n}}}＜2^{2^{\mathfrak{n}}}\cdot2$. As a consequence, $\mathfrak{n}$ is finite if and only if $2^{2^{2^{\mathfrak{n}}}}＜(2^{2^{2^{\mathfrak{n}}}})^2$. It is still open (asked by Läuchli) whether $\mathfrak{n}$ is finite if and only if $2^{2^{\mathfrak{n}}}＜(2^{2^{\mathfrak{n}}})^2$.

Läuchli, H., Ein Beitrag zur Kardinalzahlarithmetik ohne Auswahlaxiom, Z. Math. Logik Grundlagen Math. 7, 141-145 (1961). ZBL0114.01005.

For an English translation of Läuchli's paper, see here.