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),

$\kappa <\lambda ⟹{\kappa }^{\mu }<{\lambda }^{\mu }\mathrm{.}$

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, $2^{\kappa }=\mid \mathcal{P}(\kappa )\mid$, and Cantor's theorem in cardinal form: $\kappa <2^{\kappa }$). It fails already at $\kappa ={\mathrm{\aleph }}_0$, $\lambda ={\mathrm{\aleph }}_1$, $\mu ={\mathrm{\aleph }}_0$, where both powers collapse to $2^{{\mathrm{\aleph }}_0}$.

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 $\mu ={\mathrm{\aleph }}_0$. 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 ${\mathrm{\aleph }}_0^{{\mathrm{\aleph }}_0}=2^{{\mathrm{\aleph }}_0}$, so raising ${\mathrm{\aleph }}_0$ 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 $\mid {\mathbb{R}}^{\mathbb{R}}\mid$.