How to obtain the identity $\sum_k\frac{(m-k)!}{(n-k)!} = \frac{(m+1-k)!}{(m+1-n)(n-k)!}$ Hereby $$\sum_k\frac{(m-k)!}{(n-k)!} = \frac{(m+1-k)!}{(m+1-n)(n-k)!} =: SOL(k)$$
is supposed to be similar to the way of writing antiderivatives. In complete form it should be like this:
$$\sum_{k=a}^b \frac{(m-k)!}{(n-k)!} = SOL(b)-SOL(a)$$
I'm not completely sure whether it actually holds for all $a,b$. 
Okay, using Pascal's identity, I get:
$$\sum_{k=a}^b \binom{m-k}{m-n}=\\
\binom{m-a}{m-n}+\sum_{k=a+1}^b \binom{m-k}{m-n}\\
= \binom{m-a-1}{m-n} +\binom{m-a-1}{m-n-1}+\sum_{k=a+1}^b \binom{m-k}{m-n} \\
= \binom{m-a-1}{m-n} +\binom{m-a-1}{m-n-1}+\binom{m-a-1}{m-n}+\sum_{k=a+2}^b \binom{m-k}{m-n}\\
= \binom{m-a-1}{m-n-1}+2\binom{m-a-1}{m-n}+ \sum_{k=a+2}^b \binom{m-k}{m-n}\\
= \binom{m-a-1}{m-n-1}+2\binom{m-a-2}{m-n}+2\binom{m-a-2}{m-n-1} \sum_{k=a+2}^b \binom{m-k}{m-n}\\
=(\sum_{i=a+1}^{b-1} i\binom{m-a-i}{m-n-1}) + b\binom{m-b}{m-n}
$$
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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\begin{align}
&\bbox[10px,#ffd]{\sum_{k = 0}^{\infty}{\pars{m - k}! \over \pars{m - n}!}} =
\pars{m - n}!\sum_{k = 0}^{\infty}{m - k \choose  m - n}
\\[5mm] = &\
\pars{m - n}!\sum_{k = 0}^{\infty}\bracks{z^{m - n}}
\pars{1 + z}^{m - k} =
\pars{m - n}!\bracks{z^{m - n}}\pars{1 + z}^{m}\sum_{k = 0}^{\infty}
\pars{1 \over 1 + z}^{k}
\\[5mm] = &\
\pars{m - n}!\bracks{z^{m - n}}\pars{1 + z}^{m}\,
{1 \over 1 - 1/\pars{1 + z}} =
\pars{m - n}!\bracks{z^{m - n + 1}}\pars{1 + z}^{m + 1}
\\[5mm] = &\
\bbx{\pars{m - n}!{m + 1 \choose m - n + 1}}
\end{align}
A: It looks that your question is relevant to the Indefinite Sum concept.
If we have that
$$
\Delta _{\,x} \,F(x) = F(x + 1) - F(x) = f(x)
$$
then we write 
$$
F(x) = \Delta _{\,x} ^{\, - \,1} \,f(x) = \sum\nolimits_{\;x\;} {f(x)} 
$$
in particular we have
$$
F(b) - F(a) = \sum\nolimits_{\;x = \,a\,}^b {f(x)}  =  - \sum\nolimits_{\;x = \,b\,}^a {f(x)} 
$$
The relation with the standard definition of the sum of a function of discrete variable is 
$$
\sum\limits_{k = a}^{b - 1} {f(k)}  = \sum\limits_{a\, \le \,k\, < \,b} {f(k)}  = \sum\nolimits_{\;k = \,a\,}^b {f(k)}
   = F(b) - F(a)\quad \left| \matrix{ \;a,b \in \mathbb Z \hfill \cr 
  \;a \le b \hfill \cr}  \right.
$$
The first two sum are "standard", the third is meant to indicate the Indef. one.
Note the $< $ in the second ! that is exactly what is needed to allow the chain
$\sum\limits_{a\, \le \,k\, < \,b}+\sum\limits_{b\, \le \,k\, < \,c}=\sum\limits_{a\, \le \,k\, < \,c}$ with $\sum\limits_{a\, \le \,k\, < \,a}=0$ 
corresponding to
$\sum\nolimits_{\;k = \,a\,}^{\;b} +\sum\nolimits_{\;k = \,b\,}^{\;c}=\sum\nolimits_{\;k = \,a\,}^{\;c} \quad;\quad \sum\nolimits_{\;k = \,a\,}^{\;b}+ \sum\nolimits_{\;k = \,b\,}^{\;a}=0\quad \to \quad \sum\nolimits_{\;k = \,a\,}^{\;b} = - \sum\nolimits_{\;k = \,b\,}^{\;a}$ 
That premised, for the binomials we have
$$
\Delta _{\,x} \,\left( \matrix{
  x \cr 
  q \cr}  \right) = \left( \matrix{
  x + 1 \cr 
  q \cr}  \right) - \left( \matrix{
  x \cr 
  q \cr}  \right) = \left( \matrix{
  x \cr 
  q - 1 \cr}  \right)\quad  \Rightarrow \quad \sum\nolimits_{\;x\;} {\left( \matrix{
  x \cr 
  q \cr}  \right)}  = \left( \matrix{
  x \cr 
  q + 1 \cr}  \right)
$$
and therefore
$$
\eqalign{
  & \Delta _{\,x} \,\left( \matrix{
  r - x \cr 
  q \cr}  \right) = \left( \matrix{
  r - x - 1 \cr 
  q \cr}  \right) - \left( \matrix{
  r - x \cr 
  q \cr}  \right) =  - \left( \matrix{
  r - x - 1 \cr 
  q - 1 \cr}  \right) = \left( { - 1} \right)^q \left( \matrix{
  q - r + x - 1 \cr 
  q - 1 \cr}  \right)  \cr 
  & \sum\nolimits_{\;x\;} {\left( \matrix{
  r - x \cr 
  q \cr}  \right)}  =  - \left( \matrix{
  r - x + 1 \cr 
  q + 1 \cr}  \right)  \cr 
  & \sum\limits_{k = 0}^b {\left( \matrix{
  r - k \cr 
  q \cr}  \right)}  =  - \left( \matrix{
  r - \left( {b - 1} \right) + 1 \cr 
  q + 1 \cr}  \right) + \left( \matrix{
  r - 0 + 1 \cr 
  q + 1 \cr}  \right) \cr} 
$$
which applies also to complex values of $r,q,b$ if the binomial is defined through the Gamma  function.
From here you should be able to conclude for your particular case.
