How to find this binomial sum result? Jaynes (again) writes an equation that he doesn't justify, and which I don't understand.



How does he derive this? Does he use the binomial theorem? because I don't see how to use the binomial theorem here, given that there is a factor $\frac {s^n}{n!}$ in the sum that you can't take out. 
Can someone give at least a hint on how to find this?
 A: The binomial theorem is not used here.

We obtain
  \begin{align*}
\color{blue}{\sum_{n=c}^\infty}&\color{blue}{\frac{n!}{c!(n-c)!}\phi^c(1-\phi)^{n-c}\frac{\exp(-s)s^n}{n!}}\\
&=\frac{\phi^c\exp(-s)}{c!}\sum_{n=c}^\infty\frac{1}{(n-c)!}(1-\phi)^{n-c}s^n\tag{1}\\
&=\frac{\phi^c\exp(-s)}{c!}\sum_{n=0}^\infty\frac{1}{n!}(1-\phi)^{n}s^{n+c}\tag{2}\\
&=\frac{(s\phi)^c\exp(-s)}{c!}\sum_{n=0}^\infty\frac{\left(s(1-\phi)\right)^{n}}{n!}\tag{3}\\
&=\frac{(s\phi)^c\exp(-s)}{c!}\exp(s(1-\phi))\tag{4}\\
&\color{blue}{=\frac{(s\phi)^c\exp(-s\phi)}{c!}}\tag{5}\\
\end{align*}

Comment:


*

*In (1) we factor out terms independent from $n$ and cancel $n!$.

*In (2) we shift the index $n$ to start from $n=0$ and adjust within the series $n\rightarrow n+c$ accordingly.

*In (3) we factor out $s^c$ and do a little rearrangement.

*In (4) we use the exponential series expansion.

*In (5) we simplify $\exp(-s)\exp(s(1-\phi))=\exp(-s)\exp(s)\exp(-s\phi)=\exp(-s\phi)$.
A: $$\frac{(\phi s)^c \exp(-s)}{c!}\sum_{n=c}^{\infty}\frac{(s(1-\phi))^{n-c}}{(n-c)!} = \frac{\exp(-s)(\phi s)^c}{c!} \exp(s(1-\phi)) = \frac{\exp(-s\phi)(\phi s)^c}{c!}$$
In the first equality, I used the series expansion of $e^x = \sum_{n=0}^{\infty}\frac{x^n}{n!}$.
