What is the intermediate step in this negative binomial proof? On Slide 47 of these slides, there is a formula stating
$$ s_k(t) = - \sum_{j=0}^{k} s_j(0) \sum_{n=j}^{k} \frac{e^{-\lambda_n t} \prod_{m=j}^{k-1} \lambda_m }{\prod^{k}_{m=j, m \neq n} (\lambda_m - \lambda_n)} $$
I want to change it to the binomial form, and I see that the intermediate steps involve the application of the binomial theorem, I struggle to make it into a suitable form for that.
First step is the simplification with $ \lambda_k = k \lambda $ due to the problem set, so we obtain
$$ s_k(t) = - \sum_{j=0}^{k} s_j(0) \sum_{n=j}^{k} \frac{e^{- n \lambda t} \prod_{m=j}^{k-1} m }{\prod^{k}_{m=j, m \neq n} (m - n)} $$
What seems difficult is that we need to change the summation index, and even with writing out the terms I don't quite see what is going on.
I understand that the numerator can be factored out and the sign factored in.
$$ s_k(t) = \sum_{j=0}^{k} s_j(0) \prod_{m=j}^{k-1} m \sum_{n=j}^{k} \frac{- e^{- n \lambda t}  }{\prod^{k}_{m=j, m \neq n} (m - n)} $$
where the product is
$$ \prod_{m=j}^{k-1} m = \frac{(k-1)!}{(j-1)!}$$
which needs an additional $ \ (k-j)! $ in its denominator to obtain the combination outside of the summation.
The exponential outside of the sum appears due to the shift of the summation. 
$$ \sum_{n=j}^{k} e^{n \lambda t} = e^{-\lambda j t} \sum_{n=0}^{k-j} e^{-n \lambda t}  $$
I would proceed showing
$$ \frac{1}{\prod^{k}_{m=j, m \neq n} (m - n)} = \frac{(j-n)!}{(k-n)!} $$
However, the term needed outside, does not appear so I would multiply the fraction with that
$$ \frac{1}{\prod^{k}_{m=j, m \neq n} (m - n)} = \frac{(k-j)!}{(k-j)!} \frac{(j-n)!}{(k-n)!} $$
which means we could borrow it for the outside term, and be left with after the change of summation
$$  \frac{(k-j)! (j-n)!}{k-n!} \rightarrow \frac{(k-j)! (j-n-j)!}{k-n-j!} $$
where there is something wrong, because negative factorials are undefined.
Now I would evidently need to show that this is equal to 
$$ {k-j \choose n } =  \frac{(k-j)!}{n! (k-j-n)!} $$
 A: Hint: The representation $$ \frac{1}{\prod^{k}_{m=j, m \neq n} (m - n)} = \frac{(j-n)!}{(k-n)!} $$     is  not  admissible, since  $(j-n)!$ is  as   you already    noted    not defined for     negative  integral   values ($j<n$).  But,   we   can write
\begin{align*}
\frac{1}{\prod^{k}_{{m=j}\atop{m \neq n}} (m - n)}
&=\frac{1}{\prod^{n-1}_{m=j} (m - n)}\cdot\frac{1}{\prod^{k}_{m=n+1} (m - n)}\\
&=\frac{(-1)^{n-j}}{(n-j)!}\cdot\frac{1}{(k-n)!}\tag{1}
\end{align*}

Focussing on the inner sum of
  \begin{align*}
s_k(t) = -\sum_{j=0}^{k} s_j(0)  \sum_{n=j}^{k} \frac{e^{- n \lambda t} \prod_{m=j}^{k-1} m }{\prod^{k}_{m=j, m \neq n} (m - n)}\tag{2}
\end{align*}
We  obtain from (1) and (2)
  \begin{align*}
\color{blue}{\sum_{n=j}^k}\color{blue}{\frac{e^{-n\lambda  t}\prod_{m=j}^{k-1}m}{\prod_{{m=j}\atop{m\ne   n}}^k(m-n)}}
&=\sum_{n=j}^k\frac{e^{-n\lambda  t}\frac{(k-1)!}{(j-1)!}}{\prod^{n-1}_{m=j} (m - n)\prod^{k}_{m=n+1} (m - n)}\\
&=\frac{(k-1)!}{(j-1)!}\sum_{n=j}^ke^{-n\lambda  t}\frac{(-1)^{n-j}}{(n-j)!}\cdot\frac{1}{(k-n)!}\\
&=\frac{(k-1)!}{(j-1)!}\sum_{n=0}^{k-j}e^{-(n+j)\lambda  t}\frac{(-1)^{n}}{n!}\cdot\frac{1}{(k-j-n)!}\tag{3}\\
&=\binom{k-1}{j-1}e^{-j\lambda t}\sum_{n=0}^{k-j}\binom{k-j}{n}\left(-e^{-\lambda t}\right)^n\\
&\,\,\color{blue}{=\binom{k-1}{j-1}e^{-j\lambda t}\left(1-e^{-\lambda  t}\right)^{k-j}}\tag{4}
\end{align*}
  resulting in
  \begin{align*}
s_k(t)=-\sum_{j=0}^{k} s_j(0)\binom{k-1}{j-1}e^{-j\lambda t}\left(1-e^{-\lambda  t}\right)^{k-j}\tag{5}
\end{align*}

Comment:


*

*In (3) we shift the index to start with $j=0$.

*The minus sign in (2) is also part of (5) contrary to the stated formula in the paper. We check the equality of (2) and (5) for $k=1$: 
\begin{align*}
\color{blue}{s_1(t)}&= -\sum_{j=0}^{1} s_j(0) \sum_{n=j}^{1} \frac{e^{- n \lambda t} \prod_{m=j}^{0} m }{\prod^{1}_{m=j, m \neq n} (m - n)}\\
&=-s_0(0)\sum_{n=0}^{1} \frac{e^{- n \lambda t} \prod_{m=0}^{0} m }{\prod^{1}_{m=0, m \neq n} (m - n)}
-s_1(0)\sum_{n=1}^{1} \frac{e^{- n \lambda t} \prod_{m=1}^{0} m }{\prod^{1}_{m=1, m \neq n} (m - n)}\\
&=0-s_1(0)\frac{e^{-\lambda t}\cdot 1}{1}\\
&\,\,\color{blue}{=-s_1(0)e^{-\lambda t}}\\
\color{blue}{s_1(t)}&=-\sum_{j=0}^1 s_j(0)\binom{0}{j-1}e^{-j\lambda t}\left(1-e^{-\lambda t}\right)^{1-j}\\
&=-s_0(0)\cdot 0\cdot e^0\left(1-e^{-\lambda t}\right)-s_1(0)\binom{0}{0}e^{-\lambda t}\left(1-e^{-\lambda t}\right)^0\\
&\,\,\color{blue}{=-s_1(0)e^{-\lambda t}}
\end{align*}
affirming the validity of the minus sign.
Here we use the empty product $\prod_{m\in\emptyset}m=1$ and $\binom{p}{q}=0$ if $q<0$ (according for instance to (5.1) in Concrete Mathematics
 by R.L. Graham, D.E. Knuth and O. Patashnik).
