I am working through the proof that $P_{ij}(s)=F_{ij}(s)P_{jj}(s)$ for $i \neq j $ where $P_{ij}(s)$ and $F_{ij}(s)$ are generating functions
I am stuck on this step
$$\sum_{n=1}^\infty p_{ii}^{(n)} s^n =\sum_{n=1}^\infty \sum_{k=1}^n f_{ii}^{(k)} p_{ii}^{(n-k)} s^n =\sum_{k=1}^\infty f_{ii}^{(k)} s^k\sum_{n=k}^\infty p_{ii}^{(n-k)} s^{n-k} $$
How is the final equality achieved?
Note that $n$ takes values from 1 to $\infty$ and k takes values from 1 to $n$. In other words, the effective range of values that k can take is 1 to $\infty$. Further, in $p_{ii}^{(n-k)}$, the number of steps $(n-k)\geq 0$, which means the minimum value that can be set for $n$ is $k$. Now, separate the terms that contain only $k$ and interchange the summations to get the final result.