Solution
= Solution
Assertion (iii) can be false. Let $\kappa=\aleph_0$ and $\lambda=2^{\aleph_0}$. Part (ii) and the <currying law for cardinal exponentiation> give
$$
\kappa^\lambda=2^\lambda,
\qquad
\lambda^\kappa=(2^\kappa)^\kappa
=2^{\kappa^2}=2^\kappa=\lambda.
$$
By <Cantor theorem>, $2^\lambda>\lambda$, and therefore
$$
\boxed{\kappa^\lambda\ne\lambda^\kappa}.
$$