FALSE: $\kappa < \lambda$ implies $\kappa^{\mu} < \lambda^{\mu}$

FALSE. Assume the Axiom of Choice (The Axiom of Choice). Cardinal exponentiation is strictly monotone in the base: for all cardinals $\kappa, \lambda, \mu$ (Cardinal sum $\kappa \oplus \lambda$, product $\kappa \otimes \lambda$ and exponentiation $\kappa^{\lambda}$, and why they are written apart from the ordinal operations),

$\kappa < \lambda \quad \Longrightarrow \quad \kappa^{\mu} < \lambda^{\mu} .$

The claim is plausible because the weak form is true — $\kappa \le \lambda$ does give $\kappa^{\mu} \le \lambda^{\mu}$ (Commutativity, associativity, distributivity and monotonicity of $\oplus$ and $\otimes$, the unit laws, the two exponent laws, and $\kappa \le \lambda$ if and only if $\kappa$ injects into $\lambda$) — and because Cantor's theorem supplies the different strict inequality $\kappa < 2^{\kappa}$ at every cardinal (Assuming the Axiom of Choice, $2^{\kappa} = \lvert \mathcal{P}(\kappa) \rvert$, and Cantor's theorem in cardinal form: $\kappa < 2^{\kappa}$). It fails already at $\kappa = \aleph_0$, $\lambda = \aleph_1$, $\mu = \aleph_0$, where both powers collapse to $2^{\aleph_0}$.

Why the collapse happens. For an infinite exponent $\mu$: once the base is at least $2$ and at most $2^{\mu}$, raising it to the power $\mu$ gives the same value, because $(2^{\mu})^{\mu} = 2^{\mu \otimes \mu} = 2^{\mu}$ squeezes the chain shut. So for infinite $\mu$ strict monotonicity in the base can only survive where the base is allowed to exceed $2^{\mu}$, and it fails on the whole interval below it; for finite exponents $\ge 1$ the collapse does not occur, which is why the witness above takes $\mu = \aleph_0$. That is a fact about Hessenberg: $\kappa \otimes \kappa = \kappa$ for every infinite cardinal $\kappa$, proved in ZF from the canonical well-order of $\kappa \times \kappa$ as much as about exponentiation.

The weak form is not damaged. $\kappa \le \lambda \Rightarrow \kappa^{\mu} \le \lambda^{\mu}$ remains true, and so does Cantor's strict inequality $\kappa < 2^{\kappa}$. What fails is the strict form in the base, and it fails at the smallest infinite instance.

A second casualty of the same computation. The chain in step 2.1 also shows $\aleph_0^{\aleph_0} = 2^{\aleph_0}$, so raising $\aleph_0$ to its own power adds nothing beyond taking the power set of $\omega$. The companion page carries that computation on its own, together with the corresponding value for $\lvert \mathbb{R}^{\mathbb{R}} \rvert$.