= Solution
<Ordinal exponentiation> is defined by <transfinite recursion>:
$$
\alpha^0=1,\qquad
\alpha^{\beta+1}=\alpha^\beta\alpha,\qquad
\alpha^\lambda=\sup_{\xi<\lambda}\alpha^\xi
$$
when $\lambda$ is a nonzero <limit ordinal>. If $\alpha\geq1$, each successor step and each supremum show that $\beta\mapsto\alpha^\beta$ is nondecreasing. If $\alpha\geq2$, <transfinite induction> strengthens this to strict increase: $\alpha^\beta<\alpha^{\beta+1}$, and the value at a limit exceeds every earlier value.
Strict increase can disappear when the base changes. For example, with $\omega<\omega+1<\omega+2$,
$$
(\omega+1)^\omega=(\omega+2)^\omega=\omega^\omega,
$$
because for every positive finite $n$, both finite powers have leading term $\omega^n$, and taking the supremum gives $\omega^\omega$.
We next prove
$$
\boxed{\ \alpha^{\beta+\gamma}=\alpha^\beta\alpha^\gamma\ }.
$$
Apply <transfinite induction> to $\gamma$. The zero and successor cases follow directly from the recursive definitions and the associativity of <ordinal multiplication>. At a limit $\lambda$, continuity of ordinal multiplication in its right argument gives
$$
\alpha^{\beta+\lambda}
=\sup_{\xi<\lambda}\alpha^{\beta+\xi}
=\sup_{\xi<\lambda}\alpha^\beta\alpha^\xi
=\alpha^\beta\alpha^\lambda.
$$
The superficially similar identity $(\alpha\beta)^\gamma=\alpha^\gamma\beta^\gamma$ is false: for $\alpha=\omega$, $\beta=2$, and $\gamma=2$,
$$
(\omega2)^2=\omega^2 2\ne\omega^2 4=\omega^2 2^2.
$$
Finally,
$$
\boxed{\ \alpha^{\omega_1}\geq\omega_1\iff\alpha\geq2\ },
\qquad
\boxed{\ \alpha^{\omega_1}\geq\omega_2\iff\alpha\geq\omega_2\ }.
$$
For the first equivalence, bases zero and one fail, while strict increase embeds every countable ordinal into the powers of any base $\alpha\geq2$. For the second, $\alpha\geq\omega_2$ immediately gives $\alpha^{\omega_1}\geq\alpha$. Conversely, if $\alpha<\omega_2$, then $|\alpha|\leq\aleph_1$. Every exponent below the <first uncountable ordinal> $\omega_1$ is countable, so each $\alpha^\beta$ has cardinality at most $\aleph_1$; the supremum of $\aleph_1$ such ordinals still has cardinality at most $\aleph_1$ and is therefore below the <second uncountable ordinal> $\omega_2$.
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