Let $N$ being a Poisson-binomial random integer with success rates $r_1,\dots,r_K \in [0,1]$. Since $N$ can be written as the sum of $K$ independent Bernoulli integers $N_i$ (each one with success rate $r_i$), i.e. \begin{equation*} N=\sum_{i=1}^K N_i \end{equation*} it follows that the Probability Generating Function (PGF) of $N$ is factorized as \begin{equation*} G_{N}(s)=\prod_{i=1}^K G_{N_i}(s)=\prod_{i=1}^K(1-r_i+r_is) \end{equation*} where it was noted that the PGF of $N_i$ is $G_{N_i}(s)=(1-r_i+r_is)$. Now, I would like to obtain the same result by employing the definition of PGF, that is \begin{equation*} G_N(s)\triangleq \sum_{n=0}^\infty p_{N}(n)\,s^n \end{equation*} where $p_N(n)$ is the probability distribution of $N$, and can be written in the following form \begin{equation*} p_N(n)=\sum_{i_{1:n}\in P_{n,K}^{>}} \prod_{\substack{i=1\\i\neq i_{1:n}}}^{K}(1-r_i)\,\prod_{j=1}^n r_{i_j} \qquad 0\leq n \leq K \end{equation*} where are introduced the following symbols:
- $i_{1:n}=(i_1,i_2,\dots, i_j, \dots, i_n)$ is a string of $n$ indexes $i_j$
- $P_{n,K}^{>}$ is a matrix composed by all strictly increasing ($i_{j+1}>i_j$) strings $i_{1:n}$ where each index $i_j$ can get value in $\{1,2,\dots,K\}$. Each row of $P_{n,K}^{>}$ is a string $i_{1:n}$
- $\sum_{i_{1:n}\in P_{n,K}^{>}}$ means that the summation is taken over the rows of $P_{n,K}^{>}$
- $\prod_{\substack{i=1\\i\neq i_{1:n}}}^{K}$ means that the product is taken over all indexes $i\in\{1,2,\dots,K\}$ that are not used in $i_{1:n}$
In particular, $p_n(0)=\prod_{i=1}^K (1-r_i)$.
Example
Let $n=2$ and $K=3$, then \begin{equation*} P_{2,3}^{>}=\left[\begin{array}{cc} 1 & 2 \\ 1 & 3 \\ 2 & 3 \end{array}\right] \qquad \rightarrow \qquad p_N(2)=(1-r_3)\,r_1\,r_2+(1-r_2)\,r_1\,r_3+(1-r_1)\,r_2\,r_3 \end{equation*}
Question
I want to prove the following identity \begin{equation}\tag{$\star$} \sum_{n=0}^K \left(\sum_{i_{1:n}\in P_{n,K}^{>}} \prod_{\substack{i=1\\i\neq i_{1:n}}}^{K}(1-r_i)\,\prod_{j=1}^n r_{i_j}\right)\,s^n=\prod_{i=1}^K (1-r_i+r_i s) \end{equation}
A simplified problem
If the $K$ success rates $r_1,\dots,r_K$ are equal to some $r$, then the previous Poisson-binomial model reduces to the more conventional binomial model, that is \begin{equation*} G_N(s)=(1-r_i+r_i s)^K \end{equation*} and \begin{equation*} p_N(n)=\binom{K}{n}\,(1-r)^{K-n}\,r^n \end{equation*} in this case my problem reduces to prove the identity \begin{equation*} \sum_{n=0}^K \left(\binom{K}{n}\,(1-r)^{K-n}\,r^n\right) s^n = (1-r+r s)^K \end{equation*} which can be easily handled with the binomial theorem. From this point of view, I'm wondering if it exists a sort of generalized binomial theorem that can be used to prove the identity $(\star)$.