[proofplan]
We write $G_{X+Y}(z) = \mathbb{E}[z^{X+Y}]$, factor the exponential as $z^X \cdot z^Y$, and use the independence of $X$ and $Y$ to split the expectation into $\mathbb{E}[z^X] \cdot \mathbb{E}[z^Y] = G_X(z) \cdot G_Y(z)$.
[/proofplan]
[step:Express $G_{X+Y}(z)$ as $\mathbb{E}[z^{X+Y}]$ and factor]
By definition of the probability generating function,
\begin{align*}
G_{X+Y}(z) = \mathbb{E}[z^{X+Y}] = \mathbb{E}[z^X \cdot z^Y].
\end{align*}
[/step]
[step:Apply independence to factorise the expectation]
Since $X$ and $Y$ are independent and both take values in $\{0, 1, 2, \ldots\}$, the random variables $z^X$ and $z^Y$ are functions of independent random variables and are therefore independent. For $z \in [0, 1]$, both $z^X$ and $z^Y$ are bounded non-negative random variables, so the expectation of their product equals the product of their expectations:
\begin{align*}
\mathbb{E}[z^X \cdot z^Y] = \mathbb{E}[z^X] \cdot \mathbb{E}[z^Y] = G_X(z) \cdot G_Y(z).
\end{align*}
Combining the two displayed equalities gives $G_{X+Y}(z) = G_X(z) \cdot G_Y(z)$.
[/step]