You can embed any countable ordinal in the rational numbers. I think this can get you interesting results in the derived set of any countable ordinal depth; for instance, a set of order type $\omega^\omega$ should do what you want.
To construct one, start with an $\omega$ between 0 and 1 (take, say, $\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \ldots$). Now add an $\omega^2$ between 1 and 2 by squeezing what you did in the first interval between 1 and $1\frac{1}{2}$, then again between $1\frac{1}{2}$ and $1\frac{2}{3}$, etc. Next make an $\omega^3$ in the next interval... you get the point. This gives you a set of order type $\omega^\omega$ but it stretches to infinity - you can squash it all though to reside in a finite interval, and this should get you a new accumulation point at step $\omega+1$.
EDIT: Here's a more explicit version. Let $S_0=\{1\}$ and think of $S_0$ as residing in the half-open interval $(0,1]$. Inductively, suppose you have $S_{n-1}$ constructed in the interval $(n-1,n]$. Construct $S_n$ by linearly mapping the interval $(n-1,n]$ onto the interval $\left(n, n+\frac{1}{2}\right]$ and taking the image of $S_{n-1}$, then again map that interval onto $\left(n+\frac{1}{2}, n+\frac{2}{3}\right]$ and take the image of $S_{n-1}$, and so on $\omega$ times mapping the same set into the intervals $\left(n+\frac{k}{k+1}, n+\frac{k+1}{k+2}\right]$. This gives you $S_n$; now let $S = \cup S_n$ be the union of all those sets.
For example, $S_1 = \{1\frac{1}{2}, 1\frac{2}{3}, 1\frac{3}{4}, \ldots\}$. $S_2$ is made up of infinitely many sequences converging on the points $2\frac{1}{2}, 2\frac{2}{3}$ etc.
You'll observe that $S_0$ has order type $1$, $S_1$ has order type $\omega$, $S_2$ has order type $\omega^2$ and generally $S_n$ has order type $\omega^n$. $S$ therefore has order type $\omega^\omega$.
You'll also observe that S_0' = \emptyset while S_1' = S_0+1 (a copy of $S_0$ shifted 1 to the right), S_2' = S_1+1 since $S_2$ has accumulation points exactly where $S_1$ has elements (shifted 1 to the right, of course), and so on and so forth. Thus $S^m$ is non-empty for all natural $m$ (in fact, $S_m$ is just $S+m$), however $S^\omega$ is empty.
To remedy that, map the half-line $[0,\infty)$ monotonically onto a finite interval such as $[0,1)$ (e.g. let $t \mapsto \frac{t}{t+1}$) and let $P$ be the image of $S$ under that mapping. $P$ has an accumulation point at $1$, and that accumulation point remains also an accumulation point of P', P'' and generally $P^m$ for any natural $m$. Therefore $P^\omega=\{1\}$ while $P^{\omega+1}=\emptyset$ as desired.