In Feller's Itroduction to Probability Teory it was written that when an interval is partitioned into two non-overlapping intervals their contributions are stochastically independent and add to $S_N$. Where $S_N=X_1+...+X_n$.
This means that: $h_{t+r}(s)=h_t(s)h_r(s)$. Every compound Poisson generating function $h_t(s)=e^{-\lambda t+ \lambda tf(s)}$ satisfies it. Where $f(s)$ is the generating function of $X_i$'s that have common distribution $f_i$. $h_i=e^{\lambda t} \sum (\lambda t)^n (f_i)^{n*}/n!$
Proof: A probability generating function $h$ is called infinitely divisible if for each positive integer $n$ the $n$-th root $h^{1/n}$ is again a probability generating function. The explanation is that if a family of probability generating functions satisfy $h_{t+r}(s)=h_t(s)h_r(s)$ then $h_t^{1/n}=h_{t/n}$ and so it is infinitely divisible. I see why $h_t^{1/n}=h_{t/n}$ but I don't see why it is sufficient to say that generating function is infinitely divisible.
Continuation
Theorem: The only infinitely divisible probability generating functions are those of the form $h_t(s)=e^{-\lambda t+ \lambda tf(s)}$ with $f_j$ a probability distribution on $0, 1, ....$
Proof: Put $h(s)=\sum h_ks^k$, and suppose that for $n \ge1$, $h^{1/n}$ is probability generating function. It follows that $h^{1/n}(s) ->1$ for every $0\le s \le 1$.
$log (\frac {h(s)}{h_0})^{1/n}=log (1+\frac {h(s)}{h_0}-1)^{1/n}$ ~$(\frac {h(s)}{h_0)})^{1/n}-1$
From where he takes log and why? Why $h^{1/n}(s)$ goes to 1?
