Question about branching process in Durrett Q) Let $Z_n$ be a branching process with offspring distribution $p_k$ i.e. $Z_{n+1} = \xi_{1}^{n+1}+...+\xi_{Z_n}^{n+1}$ if $Z_n>0$ and $Z_{n+1} = 0$ if $Z_n = 0$ where $\xi_i^n$ are i.i.d and $p_k = P(\xi_i^n = k)$ is the offspring distribution. Let $\phi(\theta) = \sum p_k\theta^k$. Suppose $\rho<1$ has $\phi(\rho)=\rho$. Show that $\rho^{Z_n}$ is a martingale and conclude $P(Z_n = 0 \text{ for some } n\geq 1|Z_0 = x)=\rho^x$. 
I've shown $\rho^{Z_n}$ is a martingale but am not sure how to conclude $P(Z_n = 0 \text{ for some } n\geq 1|Z_0 = x)=\rho^x$ because $E(\rho^{Z_n}|Z_0 = x) = \rho^x$ and the expectation is for a fixed $n$ where as the required probability is for some $n$? Thanks.
 A: Note that $Z_n=0$ implies $Z_{n+1}=0$. If $A_n$ is an increasing sequence of events with $P(A_n)=t$ for all $n$ then $P(\cup_n A_n)=\lim_n P(A_n)=t$. 
A: Here, you can prove that $$\{\lim Z_n / \mu^n > 0 \} = \{Z_n > 0 \mbox{ for all } n\} a.s.$$
With this note, let $N = \inf \{ n : Z_n = 0\}$ be a stopping time. Since $\rho^{Z_n}$ is martingale, $\rho^{Z_{n \wedge N}}$ is martingale, and thus $$\rho^x = \mathbb{E}[\rho^{Z_{0 \wedge N}}] = \mathbb{E}[\rho^{Z_{n \wedge N}}].$$ Since $\rho < 1$, we can apply the dominated convergence theorem, to get $$\rho^x = \rho^0 \mathbb{P}(N < \infty) + \rho^{\infty}\mathbb{P}(N = \infty) = \mathbb{P}(N < \infty)$$ where the first equality from the first notification that I gave above.

EDIT:
Above almost sure equality may hold under the condition that $\mathbb{P}(\lim Z_{n}/\mu^n = 0) < 1$ and I'm not sure whether it holds or not.
First, we may assume that $p_0 > 0$ since otherwise it becomes trivial.
Now, the following lemma is helpful:

Lemma let $X_n$'s be random variables taking values in $[0, \infty)$, and $D = \{ X_n = 0 \mbox{ for some } n \ge 1\}$. Assume that $\mathbb{P}(D | X_1, \cdots, X_n) \ge \delta(x) > 0$ a.s. on $\{X_n \le x\}$, $$\mathbb{P}(D \cup \{\lim X_n = \infty\}) = 1$$
Proof On $\{\liminf X_n \le M \}$, $X_n \le < M + 1$ infinitely often so $$\mathbb{P}(D | X_1, \cdots, X_n) \ge \delta(M+1) > 0$$ i.o.
By Levy's 0-1 law, LHS converges to $1_D$ so we have $\{\liminf X_n \le M \} \subseteq D$. Taking $M \to \infty$, $\{\liminf X_n < \infty \} \subseteq D$ a.s. and the result follows.

Now, from $p_0 > 0$, $\mathbb{P}(Z_{n+1} = 0 | Z_1, \cdots, Z_n) \ge p_0 ^k$ on $\{Z_n \le k\}$. Thus by the lemma, $$\mathbb{P}(\{Z_n = 0 \mbox{ for some } n\} \cup \{\lim Z_n = \infty\}) = 1$$ which allows us to do the calculation in the original answer.
