Exercise 19.7 of Jech's Set Theory says:
Assume $V = L[D]$. If $U$ is a $\kappa$-complete nonprincipal ultrafilter on $\kappa$ and if $U \neq D$, then there is a monotone function $f : \kappa \to \kappa$ such that $\kappa \leq [f]_U < [d]_U$. (Hence $U$ does not extend the closed unbounded filter.)
The hint below says:
$U$ satisfies (19.15) for some $\delta$; if $\delta = \kappa^{(n)}$; if $\delta = \kappa^{(n)}$ for some $n$, then $U = D$. Let $n$ be such that $\kappa^{(n-1)} < \delta < \kappa^{(n)}$; let $g : \kappa^n \to \kappa$ represents $\delta$ in $\operatorname{Ult}_{D_n}$. Let $f(\xi) = $ least $\alpha$ such that $g(\alpha_1,\dots,\alpha_{n-1},\alpha) \geq \xi$ for some $\alpha_1 < \ldots < \alpha_{n-1} < \alpha$. The function $f$ is monotone. To show that $[f]_U < [d]_U$, we argue as follows: For almost all (mod $D_n$) $\alpha_1,\dots,\alpha_n$, $g(\alpha_1,\dots,\alpha_n) > \alpha_n$ hence for almost all $\alpha_1,\dots,\alpha_n$, $f(g(\alpha_1,\dots,\alpha_n)) < g(\alpha_1,\dots,\alpha_n)$. Hence $(j_{D_n}(f))(\delta) < \delta$, and hence for almost all $\xi$ (mod $U$), $f(\xi) < \xi$. Thus $[f]_U < [d]_U$.
Here, "$U$ satisfies (19.15)" means that $U = \{X \subseteq \kappa : \delta \in i_{0,\omega}(X)\}$ for some $\delta < \kappa^{(\omega)}$. This is Lemma 19.21.
I would like to understand the first line of the hint, which says that "if $\delta = \kappa^{(n)}$ for some $n$, then $U = D$". I'm trying to figure it out, and I suspect I am missing something obvious but I cannot seem to see why we must have $U = D$. So far, I have that: \begin{align*} \kappa^{(n)} \in i_{0,\omega}(X) &\iff i_{n+1,\omega}(\kappa^{(n)}) \in i_{0,\omega}(X) \\ &\iff i_{n+1,\omega}(\kappa^{(n)}) \in i_{n+1,\omega}(i_{0,n+1}(X)) \\ &\iff \kappa^{(n)} \in i_{0,n+1}(X) \end{align*} so we have $U = \{X \subseteq \kappa : \kappa^{(n)} \in i_{0,n+1}(X)\}$. Since $D = \{X \subseteq \kappa : \kappa \in i_{0,1}(X)\}$, I thought that perhaps I can "pull back" the sentence $\kappa^{(n)} \in i_{0,n+1}(X)$ to $\kappa \in i_{0,1}(X)$, but I can't see how this can be done.
What we want to show is $i_{0,n}(\kappa)\in j^{(n)}(i_{0,n}(X))$ if and only if $\kappa\in j^{(0)}(X)$, where $j^{(k)}=i_{k,k+1}\colon \mathrm{Ult}^{(k)}\to \mathrm{Ult}^{(k+1)}$.
We can see that $j^{(0)}$ is the elementary embedding given by the ultrapower from $D$. Similarly, $\mathrm{Ult}^{(n)}$ thinks $j^{(n)}$ is the elementary embedding given by the ultrapower from $i_{0,n}(D)$.
Hence if we let $\phi(\kappa, D, X)$ be $\kappa\in j^{(0)}(X)$, then $$\mathrm{Ult}^{(n)}\models \phi(i_{0,n}(\kappa),i_{0,n}(D),i_{0,n}(X))$$ must be $i_{0,n}(\kappa)\in j^{(n)}(i_{0,n}(X))$.