Simplification of an expression. I've been working on analysing one system lately, I arrived to an expression which I'd like to simplify more, still couldn't manage to.
I arrived to the following expression :
$\sum_{k = n+2}^{N+1}\gamma^k\prod_{t=n+2}^{k}(1-(t-n-1)\alpha); n \in N , \alpha \in R$, and $\gamma<1$.
I just included the sum operator for illustration.
What I'd like to know, if there's a way to write :
$\prod_{t=n+2}^{k}(1-(t-n-1)\alpha)$
in a more simplified way. I was thinking about the Gamma function, since this expression looks like a combination of factorials somehow to me, but didn't manage to figure out how can I arrive there.
Thank you
 A: We can employ the Gamma function to get a convenient representation.

We obtain for $\alpha\in\mathbb{C}\setminus\mathbb{Z}$:
\begin{align*}
\color{blue}{\prod_{t=n+2}^{k}}&\color{blue}{\left(1-\left(t-n-1\right)\alpha\right)}\\
&=\prod_{t=1}^{k-n-1}\left(1-t\alpha\right)\tag{1}\\
&=\alpha^{k-n-1}\prod_{t=1}^{k-n-1}\left(\frac{1}{\alpha}-t\right)\tag{2}\\
&=\alpha^{k-n-1}\binom{\frac{1}{\alpha}-1}{k-n-1}(k-n-1)!\tag{3}\\
&=\alpha^{k-n-1}\frac{\Gamma\left(\frac{1}{\alpha}\right)}{\Gamma\left(\frac{1}{\alpha}-k+n+1\right)\Gamma(k-n)}(k-n-1)!\tag{4}\\
&\,\,\color{blue}{=\frac{\alpha^{k-n-1}\Gamma\left(\frac{1}{\alpha}\right)}{\Gamma\left(\frac{1}{\alpha}-k+n+1\right)}}
\end{align*}

Comment:

*

*In (1) we shift the index by $n+1$ to start with $t=1$.


*In (2) we factor out $\alpha^{k-n-1}$.


*In (3) we use the definition of binomial coefficients for $\beta \in \mathbb{C}, q\in \mathbb{N}$
\begin{align*}
\binom{\beta}{q}=\frac{\beta(\beta-1)\cdots(\beta-q+1)}{q!}
\end{align*}


*In (4) we use the identity $\binom{\beta}{q}=\frac{\Gamma\left(\beta+1\right)}{\Gamma\left(\beta - q + 1\right)\Gamma\left(q+1\right)}$.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[5px,#ffd]{\left. \sum_{k\ =\ n + 2}^{N + 1}\gamma^{k}
\prod_{t\ =\ n + 2}^{k}\bracks{1 -\pars{t - n - 1}\alpha}
\,\right\vert_{{\Large\ n\ \in\ \mathbb{N}} \atop
{{\Large\ \alpha\ \in\ \mathbb{R}} \atop{\Large\ \gamma\ <\ 1}}}}
\\[5mm] = &\
\sum_{k\ =\ n + 2}^{N + 1}\gamma^{k}\,\pars{-1}^{k - n - 1}\,
\alpha^{k - n - 1}
\prod_{t\ =\ n + 2}^{k}\pars{t - n - 1 - {1 \over \alpha}}
\\[5mm] = &\
\pars{-\alpha}^{-n - 1}\sum_{k\ =\ n + 2}^{N + 1}\pars{-\alpha\gamma}^{k}
\pars{1 - {1 \over \alpha}}^{\overline{k - n - 1}}
\\[5mm] = &\
\pars{-\alpha}^{-n - 1}\sum_{k\ =\ n + 2}^{N + 1}\pars{-\alpha\gamma}^{k}\,
{\Gamma\pars{k - n - 1/\alpha} \over \Gamma\pars{1 - 1/\alpha}}
\\[5mm] = &\
\bbx{{1 \over \pars{-\alpha}^{n + 1}\,\Gamma\pars{1 - 1/\alpha}}\,\,
\sum_{k\ =\ n + 2}^{N + 1}\pars{-\alpha\gamma}^{k}\,
\pars{k - n - 1 - {1 \over \alpha}}!} \\ &
\end{align}
