= Solution
Let $G(s)=\mathbb E[s^\xi]$ be the offspring <probability generating function>, and let $G_n(s)=\mathbb E[s^{Z_n}]$, for $0\leq s\leq1$. Given $Z_n=k$, the next generation is a sum of $k$ <independent> offspring counts. Multiplication of their <probability generating functions> gives
$$
\mathbb E[s^{Z_{n+1}}\mid Z_n=k]=G(s)^k.
$$
Taking <conditional expectation> and then averaging proves
$$
\boxed{G_{n+1}(s)=G_n(G(s)),\qquad G_0(s)=s.}
$$
Thus $G_n$ is the $n$-fold composition of $G$ with itself. Induction makes this precise: it is true for $n=0$, and composing once more gives the next generation. In particular $G_n\circ G=G\circ G_n$, since both are the same iterated function. This is the generating-function recursion for a <Galton-Watson process>.
Write $m=\mathbb E Z_1=G'(1^-)$, initially assuming it is finite. The <law of total expectation> gives
$$
\mathbb E[Z_{n+1}\mid Z_n]=mZ_n,
\qquad \mathbb E Z_{n+1}=m\mathbb E Z_n.
$$
Starting at one therefore yields $\boxed{\mathbb E Z_n=m^n}$, with the value at $n=0$ understood to be one even if $m=0$. Equivalently, differentiating the composition at one gives the same recurrence. If the offspring <expectation> is infinite, every positive generation has infinite <expectation>: a positive chance of at least one parent remains at each finite generation, and conditional on any positive parent count the next expected count is infinite.
For a <Poisson branching process>, $G(s)=e^{\lambda(s-1)}$. Extinction is absorbing because an empty generation has no parents, so extinction by generation $n$ is precisely $Z_n=0$. Hence $x_n=G_n(0)$. Using the composition in the order $G\circ G_n$ gives
$$
\boxed{x_{n+1}=e^{\lambda(x_n-1)},\qquad x_0=0.}
$$
For example $x_1=e^{-\lambda}$ is the chance of no offspring in the first generation. The recurrence includes the degenerate case $\lambda=0$, where extinction occurs in the first generation with certainty.
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