Past exam of the logic and set theory course of the University of Oxford / 2010 / B / 4 / e

False. Take $\kappa =2^{{\aleph }_0}$ and $\lambda ={\aleph }_0$, both infinite. The explicit pairing bijection between ${\mathbb{N}}^2$ and $\mathbb{N}$ and the currying law for cardinal exponentiation giveThe power-set cardinal need not be assumed well-orderable. Every bijection used in this calculation is explicitly available in ZF, so the counterexample does not use choice.