[proofplan] We prove the recurrence $F_{n+1}(z) = F_n(F(z))$ by induction on $n$. The base case $F_1(z) = F(z)$ is immediate from the definition. For the inductive step, we condition on the first generation: given that $Z_1 = k$, the population at generation $n+1$ is the sum of $k$ independent copies of $Z_n$, so $F_{n+1}(z) = F_1(F_n(z)) = F(F_n(z))$. The inductive hypothesis $F_n(z) = F \circ \cdots \circ F(z)$ ($n$ times) then gives $F_{n+1}(z) = F_n(F(z))$. [/proofplan] [step:Establish the base case $F_1(z) = F(z)$] Let $F(z) = G_{Z_1}(z) = \mathbb{E}[z^{Z_1}]$ be the offspring pgf, and let $F_n(z) = G_{Z_n}(z) = \mathbb{E}[z^{Z_n}]$ be the pgf of the $n$-th generation population size, starting from $Z_0 = 1$. For $n = 1$, we have $F_1(z) = \mathbb{E}[z^{Z_1}] = F(z)$, which is $F$ composed once. [/step] [step:Condition on the first generation to relate $F_{n+1}$ to $F$ and $F_n$] The population at generation $n+1$ satisfies \begin{align*} Z_{n+1} = \sum_{i=1}^{Z_1} Z_n^{(i)}, \end{align*} where, given $Z_1 = k$, the random variables $Z_n^{(1)}, \ldots, Z_n^{(k)}$ are independent copies of $Z_n$ (each representing the descendants after $n$ further generations of one individual in the first generation). Conditioning on $Z_1$, \begin{align*} F_{n+1}(z) = \mathbb{E}[z^{Z_{n+1}}] = \sum_{k=0}^{\infty} \mathbb{E}\bigl[z^{Z_n^{(1)} + \cdots + Z_n^{(k)}}\bigr]\, \mathbb{P}(Z_1 = k). \end{align*} Since $Z_n^{(1)}, \ldots, Z_n^{(k)}$ are independent, each with pgf $F_n$, \begin{align*} \mathbb{E}\bigl[z^{Z_n^{(1)} + \cdots + Z_n^{(k)}}\bigr] = (F_n(z))^k. \end{align*} Substituting back, \begin{align*} F_{n+1}(z) = \sum_{k=0}^{\infty} (F_n(z))^k\, \mathbb{P}(Z_1 = k) = F(F_n(z)). \end{align*} This is precisely the pgf of $Z_1$ evaluated at $w = F_n(z)$. [/step] [step:Complete the induction to obtain $F_{n+1}(z) = F_n(F(z))$] We have shown $F_{n+1}(z) = F(F_n(z))$ for all $n \ge 0$. We now verify that this equals $F_n(F(z))$, i.e., that $F \circ F_n = F_n \circ F$ ($n+1$-fold composition can be grouped either way). We prove by induction that $F_n$ is the $n$-fold composition $F^{\circ n} := \underbrace{F \circ F \circ \cdots \circ F}_{n}$. **Base case:** $F_1 = F = F^{\circ 1}$. **Inductive step:** Assume $F_n = F^{\circ n}$. Then \begin{align*} F_{n+1}(z) = F(F_n(z)) = F(F^{\circ n}(z)) = F^{\circ (n+1)}(z). \end{align*} Since composition of a fixed function $F$ is associative, \begin{align*} F^{\circ (n+1)}(z) = F^{\circ n}(F(z)) = F_n(F(z)). \end{align*} Therefore $F_{n+1}(z) = F(F_n(z)) = F_n(F(z))$, confirming both claimed identities. [/step]