A detail on the proof of Theorem 6.9 of Steel's note on iterated ultrapowers I am reading Steel's note on iterated ultrapowers, and I am filling the gaps in the proof of Theorem 6.9. However, I am stuck on how to proceed with it.
I am checking the last part of the proof (page 31 of the linked note), which claims $P_{n+1}\in V^Q_{\operatorname{lh}E}=V^{P_n}_{\operatorname{lh}E}$. However, I do not see why the last equality holds. I tried to prove it from the equality $V^{N_n}_{\operatorname{lh}F_n}=V^{N_{n+1}}_{\operatorname{lh}F_n}=V^{\operatorname{Ult}(N_k,F_n)}_{\operatorname{lh}F_n}$. While $\pi_n(V^{N_n}_{\operatorname{lh}F_n})=V_{\operatorname{lh}E}^{P_n}$ seems obviously true, I have no idea why $\pi_n(V^{\operatorname{Ult}(N_k,F_n)}_{\operatorname{lh}F_n})=V^Q_{\operatorname{lh}E}$ holds.
I do not see how iterations over a tree work, so I must be missing something on very simple facts. I would appreciate any help.

I get how to proceed. Note that $k$ is the least number such that $\operatorname{crit} F_n<\operatorname{lh} F_k$. Moreover, if we take $\tau_k=\sup \sigma^" \operatorname{lh} F_k$, then $P_k\underset{\tau_k}{\backsim} P_n$ by the inductive assumption.
Fix $\xi$ such that $\operatorname{crit} F_n<\xi<\operatorname{lh} F_k$. Then $$\operatorname{crit}E=\pi_n(\operatorname{crit}F_n)<\pi_n(\xi)=\sigma(\xi)<\tau_k,$$ where the equality follows from Exercise 34. By Lemma 6.2, we have $\operatorname{Ult}(P_k,E)\backsim_{i_E(\operatorname{crit}E)+1} \operatorname{Ult}(P_n,E)$.
Since $\operatorname{lh}E\le i_E(\operatorname{crit}(E))$ and $P_n\models \text{$E$ is nice}$, we have $V^Q_{\operatorname{lh}E}=V^{\operatorname{Ult}(P_k,E)}_{\operatorname{lh}E}=V^{\operatorname{Ult}(P_n,E)}_{\operatorname{lh}E}=V^{P_n}_{\operatorname{lh}E}$.
