= Solution
The <Hartogs theorem> says that for every set $X$ there is an ordinal that does not inject into $X$. Indeed, the well-orderings of subsets of $X$ form a set, since each is a relation in $\mathcal P(X\times X)$. By replacement, their order types form a set of ordinals. Its supremum plus one cannot be the order type of a well-ordered subset of $X$, and hence cannot inject into $X$. The least such ordinal is the Hartogs ordinal $h(X)$.
<Ordinal exponentiation> is defined by
$$
\alpha^0=1,\qquad
\alpha^{\beta+1}=\alpha^\beta\alpha,\qquad
\alpha^\lambda=\sup_{\xi<\lambda}\alpha^\xi
$$
for nonzero <limit> $\lambda$. Fix $\alpha,\beta$ and induct on $\gamma$. The zero case is immediate. If the identity holds at $\gamma$, then
$$
\alpha^{\beta+(\gamma+1)}
=\alpha^{(\beta+\gamma)+1}
=\alpha^{\beta+\gamma}\alpha
=\alpha^\beta\alpha^{\gamma+1}.
$$
At a <limit> $\lambda$, continuity of exponentiation and right continuity of ordinal multiplication give
$$
\alpha^{\beta+\lambda}
=\sup_{\gamma<\lambda}\alpha^{\beta+\gamma}
=\sup_{\gamma<\lambda}\alpha^\beta\alpha^\gamma
=\alpha^\beta\alpha^\lambda.
$$
Put $\rho=\omega^\alpha$. The set
$$
B=\{\eta:\rho\eta\leq\gamma\}
$$
is nonempty and bounded, so let $\beta=\sup B$. Continuity of multiplication in its right argument gives $\rho\beta\leq\gamma$. There is a unique tail $\delta$ with $\gamma=\rho\beta+\delta$. If $\delta\geq\rho$, then
$\rho(\beta+1)=\rho\beta+\rho\leq\gamma$, contrary to maximality. Hence $\delta<\rho$. If
$\rho\beta+\delta=\rho\beta'+\delta'$ with both remainders below $\rho$ and $\beta<\beta'$, then
$$
\rho\beta+\delta<\rho(\beta+1)\leq\rho\beta',
$$
a contradiction. This proves existence and uniqueness in the <division by an additively indecomposable ordinal>.
Now let $X$ be nonempty and well ordered. Its least element is not a <limit> point, so $X'\ne X$. Every <derivative> is itself a well-ordered subset. If $X^{(\alpha)}$ were nonempty for every $\alpha<h(X)$, then each difference
$$
X^{(\alpha)}\setminus X^{(\alpha+1)}
$$
would be nonempty. Choosing its least member would inject $h(X)$ into $X$, since the nested successive differences are disjoint. This contradicts Hartogs' lemma. Thus some <derivative> is empty, as asserted by the <derived-set iteration of a well-order>.
For an ordinal $\xi$, a point $\gamma<\xi$ is a <limit> point exactly when it is a nonzero <limit> ordinal. Division by $\omega$ writes $\gamma=\omega\beta+n$ with $n<\omega$; this is a nonzero <limit> exactly when $n=0$ and $\beta>0$. Hence
$$
\boxed{\xi'=\{\omega\beta<\xi:\beta>0\}}.
$$
Within this derived set, $\omega\beta$ is a <limit> point exactly when $\beta$ is a nonzero <limit> ordinal, equivalently $\beta=\omega\eta$ for some $\eta>0$. Therefore the <derived sets of an ordinal> satisfy
$$
\boxed{\xi''=\{\omega^2\eta<\xi:\eta>0\}}.
$$
Finally $\omega'=\varnothing$, so the index of $\omega$ is $1$, while
$(\omega^2)'=\{\omega,2\omega,3\omega,\ldots\}$ and
$(\omega^2)''=\varnothing$, so the index of $\omega^2$ is $2$.
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