Closed expression for $\prod_{i=0}^k(1-ix)$ How do I find a closed form expression for the product $\displaystyle{\prod_{i=0}^{k-1}(1-(k-i)x)}\;$?
Edit 1: @Kenta S has found a different way to index it, do you have any other ideas yet?
 A: We can write the product using Stirling numbers of the first kind $\begin{bmatrix}k\\q\end{bmatrix}$.
\begin{align*}
\prod_{j=1}^k\left(1-jx\right)=\sum_{q=0}^k(-1)^{q}\begin{bmatrix}k+1\\k+1-q\end{bmatrix} x^q
\end{align*}
where we start with index $j=1$, since the factor with $j=0$ is $\left(1-0\cdot x\right)=1$.

Somewhat more elaborated we can write the product using $[k]=\{1,2,\ldots,k\}$ as
\begin{align*}
\color{blue}{\prod_{j=1}^k\left(1-jx\right)}
&=\sum_{S\subseteq[k]}\prod_{s\in S}\left(-sx\right)\tag{1.1}\\
&=\sum_{q=0}^k\left(\sum_{{S\subseteq[k]}\atop{|S|=q}}(-1)^{|S|}\left(\prod_{s\in S}s\right)\right)x^q\tag{1.2}\\
&\,\,\color{blue}{=\sum_{q=0}^k(-1)^{q}\begin{bmatrix}k+1\\k+1-q\end{bmatrix} x^q}\tag{1.3}
\end{align*}

Comment:

*

*In (1.1) we write the product as sum over all subsets in $[k]$. For each index $s\in S$ we select $-sx$ from the product and $1$ otherwise.


*In (1.2) we order the summation according to the size of $S$.


*In (1.3) we use the  Stirling numbers of the first kind.
Example: $k=3$
Let's look at a small example $k=3$. We obtain
\begin{align*}
\color{blue}{\prod_{j=1}^3(1-jx)}
&=(-1)^{|\emptyset|}\cdot 1\\
&\quad+\left((-1)^{|\{1\}|}\cdot 1+(-1)^{|\{2\}|}\cdot 2+(-1)^{|\{3\}|}\cdot 3\right)x\\
&\quad+\left(-1)^{|\{1,2\}|}\cdot 1\cdot 2+(-1)^{|\{1,3\}|}\cdot 1\cdot 3+(-1)^{|\{2,3\}|}\cdot 2\cdot 3\right)x^2\\
&\quad+\left((-1)^{|\{1,2,3\}|}\cdot 1\cdot 2\cdot 3\right) x^3\\
&=1-(1+2+3)x+(1\cdot 2+1\cdot 3+2\cdot 3)x^2-(1\cdot 2\cdot 3)x^3\\
&\,\,\color{blue}{=1-6x+11x^2-6x^3}\\
&\,\,\color{blue}{=\sum_{q=0}^3(-1)^{q}\begin{bmatrix}4\\4-q\end{bmatrix} x^q}
\end{align*}
