Assuming the Axiom of Choice, $2^{\kappa }=\mid \mathcal{P}(\kappa )\mid$, and Cantor's theorem in cardinal form: $\kappa <2^{\kappa }$

Assume the Axiom of Choice (The Axiom of Choice), so that every set has a cardinality (The well-ordering theorem, 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). Let $\kappa$ be a cardinal (Cardinal (initial ordinal) and cardinality) and read $2=\{0,1\}$ as a cardinal. Then:

(a) $2^{\kappa }=\mid \mathcal{P}(\kappa )\mid$ (Cardinal sum $\kappa \oplus \lambda$, product $\kappa \otimes \lambda$ and exponentiation ${\kappa }^{\lambda }$, and why they are written apart from the ordinal operations), and more generally $2^{\mid A\mid }=\mid \mathcal{P}(A)\mid$ for every set $A$;

(b) $\kappa <2^{\kappa }$.

Clause (b) is Cantor's theorem: $A\prec \mathcal{P}(A)$ transcribed into cardinal arithmetic. The underlying combinatorial fact — that there is no surjection $A\to \mathcal{P}(A)$ — is a theorem of ZF and needs no choice at all; what the Axiom of Choice buys here is only the right to write $\mid \mathcal{P}(A)\mid$ and $2^{\kappa }$ as cardinals in the first place.

Why $2$ and not some other base. The characteristic function of a subset takes two values, so the power set is the function space with base $2$; that is the whole content of step 1.1, and it is why $2^{\kappa }$ rather than $\mathcal{P}$ is the object cardinal arithmetic manipulates. For infinite $\kappa$, any base $\mu$ with $2\le \mu$ gives the same value once $\mu \le 2^{\kappa }$, by monotonicity, the second exponent law and Hessenberg: $\kappa \otimes \kappa =\kappa$ for every infinite cardinal $\kappa$, proved in ZF from the canonical well-order of $\kappa \times \kappa$, and the companion page computes one such case; for finite $\kappa$ the bases genuinely differ, $3^2=9\mathrm{\ne }4=2^2$.

No fixed point. Clause (b) holds for every cardinal, so no cardinal satisfies $2^{\kappa }=\kappa$ and the hierarchy of cardinals never terminates. The corresponding statement one level up — that $\alpha \mapsto {\mathrm{\aleph }}_{\alpha }$ has no fixed point — is false, and the companion page exhibits one; the two operations behave quite differently, and it is the power operation, not the successor operation, that is unboundedly expansive.

Where the Axiom of Choice is and is not spent. Cantor's theorem: $A\prec \mathcal{P}(A)$ is choice free, and so is step 1.1. The hypothesis is used only to know that a set has a cardinality: at $\mathcal{P}(\kappa )$ and at ${}^{\kappa }2$ for clause (b), and again at $A$ and $\mathcal{P}(A)$ in the general form of clause (a). In ZF alone, $\mathcal{P}(\omega )$ may fail to be well-orderable, and then $2^{{\mathrm{\aleph }}_0}$ is not an ordinal and the inequality of clause (b) has no cardinal to compare $\kappa$ with — while the underlying statement "there is no surjection $\omega \to \mathcal{P}(\omega )$" remains a theorem.