= Solution
Choose sets $K,L,M$ of cardinalities $\kappa,\lambda,\mu$. Their <cardinal arithmetic> operations are
$$
\kappa+\lambda=|K\sqcup L|,
\qquad
\kappa\lambda=|K\times L|,
\qquad
\kappa^\lambda=|K^L|,
$$
where $K^L$ is the set of functions $L\to K$. The relation $\kappa\le\lambda$ means that there is an <injective function> $K\to L$.
The <currying law for cardinal exponentiation> follows from the explicit <bijection>
$$
(K^L)^M\longleftrightarrow K^{L\times M},
\qquad
f\longmapsto\bigl((l,m)\mapsto f(m)(l)\bigr),
$$
and proves
$$
\boxed{(\kappa^\lambda)^\mu=\kappa^{\lambda\mu}}.
$$
Moreover, $2^\kappa$ is the cardinality of the <power set> of a set of size $\kappa$, so <Cantor theorem> gives
$$
\boxed{2^\kappa>\kappa}.
$$
For completeness, identify each cardinal with its <initial ordinal>. Suppose that some infinite $\kappa$ violates $\kappa^2=\kappa$, and choose the least such cardinal. Well-order the pairs $(\alpha,\beta)\in\kappa^2$ first by $\max\{\alpha,\beta\}$ and then lexicographically. Every proper initial segment is contained in $\gamma^2$ together with finitely many boundary pieces for some $\gamma<\kappa$, and has cardinality below $\kappa$ by minimality. The resulting well-order therefore has cardinality at most $\kappa$. The reverse inequality is immediate from $\alpha\mapsto(\alpha,0)$, contradicting the choice of $\kappa$. Thus the <square of an infinite cardinal> satisfies $\kappa^2=\kappa$.
If $\nu=\max\{\kappa,\lambda\}$, monotonicity now gives
$$
\nu\le\kappa+\lambda\le\nu+\nu=\nu,
\qquad
\nu\le\kappa\lambda\le\nu^2=\nu.
$$
Hence the <sum and product of two infinite cardinals> obey
$$
\boxed{\kappa+\lambda=\kappa\lambda=\max\{\kappa,\lambda\}}.
$$
Assertion (i) can be false. For any infinite $\lambda$, take $\kappa=(2^\lambda)^+$. Then $\kappa^\lambda\ge\kappa>2^\lambda$, so
$$
\boxed{\kappa^\lambda=2^\lambda\text{ is not always true}}.
$$
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