Let $X, X_{n,k}$ for $k,n\in\mathbb{N}$ denote independent random variables with values in $\mathbb{N}_0$. $X$ is identically distributed as all $X_{n,k}$. Define $N_0:=1$ and for $n\in\mathbb{N}$ set $$ N_n:=\begin{cases}0, & \text{ if }N_{n-1}=0\\X_{n,1}+\cdots+X_{n,N_{n-1}}, & \text{ if }N_{n-1}>0.\end{cases} $$ Find conditions on the distribution of $X$ for which the probability $$ q:=P(\exists n\in\mathbb{N}: N_n=0) $$ satisfies $q=1$.
Defining the hitting time of $0$ as $H(0)=\inf\left\{n\geq 0: N_n=0\right\}$, it is $$ q=P(H(0)<\infty). $$
Now the only idea I have is to start like this: $$ P(H(0)<\infty)=\sum_{n\in\mathbb{N}_0}P(H(0)=n)\\ =\sum_n P(N_n=0|N_{n-1}>0,...,N_0=1) $$ Markov property: $$ =\sum_n P(N_n=0|N_{n-1}>0)\\ =\sum_{n}\frac{P(N_n=0,N_{n-1}>0)}{P(N_{n-1}>0)}\\ =\sum_n \frac{P(X_{n,1}+\cdots+X_{n,N_{n-1}}=0)}{P(X_{n-1,1}+\cdots+N_{n-1,N_{n-2}}>0)} $$
X identically distributed as all $X_{n,k}$:
$$ =\sum_{n}\frac{P(N_{n-1}\cdot X=0)}{P(N_{n-2}\cdot X>0)} $$
In case that this makes any sense: How do I have to continue? Give me some help, please.