The offspring distribution $X$ of a branching process is given by Poisson($\lambda$). I need to show that the survival probability $\gamma$ satisfies the following equation:
$$ \gamma = 1 - e^{-\lambda \gamma} $$
Now suppose $\kappa$ is the extinction probability, then it is the smallest solution in $[0,1]$ of the equation $$\kappa = G_X(\kappa) = \mathbb{E}[X^\kappa]$$ where $G_X$ is the probability generating function of $X$.
Since $\kappa = 1 - \gamma$ , therefore $$1- \gamma = G_X(1-\gamma)$$ $$ \space \space \space = \mathbb{E}[X^{1-\gamma}]$$ $$= \sum_{i=0}^{\infty}i^{1-\gamma} \frac{e^{-\lambda}\lambda^i}{i !}$$ This implies, $$\gamma = 1 - e^{-\lambda} \sum_{i=0}^{\infty}i^{1-\gamma} \frac{\lambda^i}{i !}$$ Now, how do I show $$e^{-\lambda \gamma} = e^{-\lambda} \sum_{i=0}^{\infty}i^{1-\gamma} \frac{\lambda^i}{i !}$$ Is my approach corret or something wrong with it?