Finite beth iteration above an infinite cardinal

In ZFC, for an infinite cardinal $\kappa$ define the relative finite beth iteration by

${\mathrm{\beth }}_0(\kappa )=\kappa ,{\mathrm{\beth }}_{n+1}(\kappa )=2^{{\mathrm{\beth }}_n(\kappa )}(n<\omega )\mathrm{.}$

The exponent is cardinal exponentiation, and ${\mathrm{\beth }}_n(\kappa {)}^{+}$ means the successor cardinal, both with the conventions of The successor cardinal ${\kappa }^{+}$, the alephs ${\mathrm{\aleph }}_{\alpha }$, the beths ${\mathrm{\beth }}_{\alpha }$, successor and limit cardinals, and the identifications ${\mathrm{\aleph }}_0=\omega$ and ${\mathrm{\aleph }}_1={\omega }_1$. This differs from the ordinary beth hierarchy, whose initial value is $\omega$; they agree when $\kappa =\omega$. In particular ${\mathrm{\beth }}_1(\kappa )=2^{\kappa }$ and ${\mathrm{\beth }}_2(\kappa )=2^{2^{\kappa }}$.

Here is a set-sized recursion justification. Let $X_0=\kappa$ and $X_{n+1}=\mathcal{P}(X_n)$. The class-function form of Transfinite recursion on $\omega$ defines this sequence of sets. Assume AC as in The Axiom of Choice to take their cardinalities. Each $X_{n+1}$ has cardinality $2^{\mathrm{\mid }{X}_n\mathrm{\mid }}$, giving exactly the displayed recurrence and its uniqueness by induction. The equivalent natural-number recursion notation is that of The recursion theorem. This argument does not treat the proper class of all cardinals as a state set.

Only finite indices occur here. The initial index zero is included; the base cardinal is infinite and hence never zero or one. No limit-stage beth operation is needed for this relative notation.