Let $(p_k)_{k \in \mathbb{N}_0}$ be a probability mass function on $\mathbb{N}_0$ such that $p_0 > 0$. Furthermore, let $q$ be the extinction probability of the branching process $(Z_t)_{t \in \mathbb{N}_0}$ such that $Z_0 = 1$ and $P(Z_1 = k) = p_k, k \in \mathbb{N}_0.$ Prove that $p'_k := q^{k-1}p_k, k \in \mathbb{N}_0$ defines another probability distribution on $\mathbb{N}_0$.
This is just the first part of the excercise, and I already solved the other parts, but I just don't see why this is another probability distribution. Obviously, if $q = 1$, then $p'_k = p_k$, and thus it would be again a probability distribution. I think we can also exclude the case $q = 0$ since $p_0 > 0$, so we'd have to consider the case $q \in (0,1)$. But then, $lim_{k \rightarrow \infty} \ q^{k-1} = 0,$ so $\sum_{k=0}^{\infty} p'_k$ shouldn't equal $1$?
Notice that $$\sum_{k=0}^\infty p'_k = \sum_{k=0}^\infty q^{k-1} \mathbb{P}(Z_1 = k) = \mathbb{E}\left[q^{Z_1-1} \right] = \frac{G(q)}{q},$$ where $G(s) = \mathbb{E}[s^{Z_1}]$ is the PGF of $Z_1$. So your question is equivalent to showing that $q$ is a fixed point of $G$.
To prove this, you can consider $q_n \colon \!= \mathbb{P}(Z_n = 0)$ and show that:
It follows that $q_{n+1} = G_{n+1}(0) = G(G_n(0)) = G(q_n)$, then letting $n \to \infty$ proves that $q = G(q)$ by continuity of $G$. See here(Proposition 1.4 and Theorem 1.7) for more details. There might be a more direct approach but I don't see one.