Question Involving Sequences of Successive Limit Ordinals This question centers around whether the following statement is true:
If $X$ is some countable ordinal and $S^X = s_1, s_2, s_3, \dots$ is a (transfinite) sequence of successive limit ordinals starting with $s_1 = \omega$, then there always exist some $s_x \in S$ (where for each $s_x$ we have $x \in X$) where $s_x \geq X$.
E.g., if $X = \omega + 1$, then $S^X = \omega, \omega \cdot 2, \omega \cdot 3, \dots, \omega^2$ where $s_2 = \omega \cdot 2 > X = \omega + 1$.
Question:  I am thinking of $X = \omega^{\omega}$+1, where the question is what does $S^{\omega^{\omega} + 1}$ equal?
$$S^{\omega^{\omega}+1} = \omega, \omega \cdot 2, \omega \cdot 3, \dots, \omega^{\omega}, \omega^{\omega} + \omega \text{ ???}$$
 A: Note that if $\alpha$ is a countable ordinal, then the $\alpha$-th limit ordinal is also countable, since the set of all countable limit ordinals is uncountable.
Therefore, your sequence is basically an initial part of the function $f:\omega_1\to\omega_1$, that sends each $\alpha\in\omega_1$ to the $\alpha$-th limit ordinal. You can prove that this function is in fact $\alpha\mapsto \omega\cdot\alpha$ (here we assume for convenience that $0$ is a limit ordinal).
Now, this function is what is called a normal function, i.e., a strictly increasing continuous function. You could prove that a normal function has fixed points $\alpha$ such that $f(\alpha)=\alpha$ (in fact it has arbitrarily large fixed points, since its set of fixed points is a closed unbounded set).
To sketch the proof: create a sequence starting with some ordinal $\alpha_0$, and let $\alpha_{n+1}=f(\alpha_n)$. Then using that $f$ is a normal function you can show that the supremum of this sequence will be a fixed point.
Now, suppose that $\alpha$ is such a fixed point, that is, $f(\alpha)=\alpha$. Then the sequence of length $\alpha$ contains the elements $f(\beta)$ such that $\beta<\alpha$, but not the element $f(\alpha)$ itself. Since $f$ is strictly increasing, $f(\beta)<f(\alpha)=\alpha$, and thus there is no such element in the sequence greater than the length of the sequence.

As for your Question:
The first of these fixed points is indeed $\omega^\omega$, since $f(\omega^\omega)=\omega\cdot\omega^\omega=\omega^\omega$. However, there is no $x\in\omega^\omega$ such that $s_x=f(x)=\omega^\omega$, thus the statement fails.
