For every set $A$ the Hartogs number $\mathrm{\aleph }(A)$ is a cardinal, and for every cardinal $\kappa$ it is the least cardinal strictly above $\kappa$; this is a theorem of ZF

Work in ZF; no choice principle is used. For a set $A$ let $\mathrm{\aleph }(A)$ be its Hartogs number (Hartogs: an ordinal that does not inject into a given set), the least ordinal (Ordinal (von Neumann)) admitting no injection into $A$. Then:

(a) $\mathrm{\aleph }(A)$ is a cardinal (Cardinal (initial ordinal) and cardinality), for every set $A$, well-orderable or not.

(b) If $\kappa$ is a cardinal then $\kappa <\mathrm{\aleph }(\kappa )$, and every cardinal $\lambda$ with $\kappa <\lambda$ satisfies $\mathrm{\aleph }(\kappa )\le \lambda$. So $\mathrm{\aleph }(\kappa )$ is the least cardinal strictly above $\kappa$.

(c) If $\kappa$ is an infinite cardinal then so is $\mathrm{\aleph }(\kappa )$.

What this supplies, and what it does not. It gives a successor operation on cardinals in ZF alone: there is always a next one, and it is definable. It says nothing about how large the next cardinal is compared with $2^{\kappa }$; that comparison is not decided by the axioms in use here.

Why claim (a) is stated for an arbitrary set. For a cardinal $\kappa$ only the special case is needed here, but the general case costs the same two lines and is exactly what Tarski: the Axiom of Choice is equivalent to the statement that $A\times A\approx A$ for every infinite set $A$, so extending Hessenberg's theorem from the alephs to arbitrary sets is exactly as strong as choice uses: there, $A$ is an arbitrary infinite set that is not known to be well-orderable, and the argument needs $\mathrm{\aleph }(A)$ to be an infinite cardinal before Hessenberg's theorem can be applied to it.

No power set is involved. Hartogs: an ordinal that does not inject into a given set builds $\mathrm{\aleph }(A)$ from the well-ordered subsets of $A$, so the successor cardinal is obtained without ever forming $\mathcal{P}(A)$ as a size. That separation is what makes the aleph hierarchy a ZF construction while the beth hierarchy is not.

The notation collides with the ordinal successor, and this page keeps them apart. ${\alpha }^{+}=\alpha \cup \{\alpha \}$ is the ordinal successor (Ordinal (von Neumann)) and is almost never a cardinal: an infinite cardinal is a limit ordinal, so $\kappa \cup \{\kappa \}$ is not one. The cardinal successor is $\mathrm{\aleph }(\kappa )$, which is much larger. Where an item on this page writes the superscript on an ordinal letter, as A set equinumerous with some ordinal has a least such ordinal, that ordinal is a cardinal, and equinumerous sets get the same one; no choice principle is used does with ${\alpha }^{+}$, it carries the published ordinal meaning $\alpha \cup \{\alpha \}$. This item writes the cardinal successor as $\mathrm{\aleph }(\kappa )$ throughout; the abbreviation ${\kappa }^{+}:=\mathrm{\aleph }(\kappa )$, on a cardinal letter only, is introduced later on this page, and the superscript on an ordinal letter keeps the ordinal meaning there too.