= Solution
\b[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> give
$$
\boxed{\kappa^\lambda=(2^{\aleph_0})^{\aleph_0}
=2^{\aleph_0\cdot\aleph_0}=2^{\aleph_0}=\kappa.}
$$
The 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.
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