Stirling number of the first kind expansion? I'm trying to expand this expression (for $1 \leq j \leq n+1$)
$$
\prod_{i=1 ; i\neq j}^{n+1} (x-i)
$$
in terms of Stirling numbers. The falling factorial $(x)_{n}$ is defined as
$$
(x)_{n}=\prod_{i=0}^{n-1} (x-i)=\sum_{i=0}^{n}(-1)^{n-i}
\begin{bmatrix}
n\\ 
i
\end{bmatrix}x^i
=\sum_{i=0}^{n}s(n,i)x^i
$$
where $s(n,i)$ is the Stirling number of the first kind.
This gives me the expression..
$$
\prod_{i=1 ; i\neq j}^{n+1} (x-i)=\frac{1}{x-j}\sum_{i=1}^{n+2}s(n+2,i)x^{i-1}
$$
Formula fixed.
Edit:
Formula is fixed, this solution works.
 A: Lets get the definition of the Stirling numbers (of the first kind) in terms of falling factorials right 
\begin{eqnarray*}
(x)_{n}=\prod_{i=0}^{\color{red}{n-1}} (x-i)=\sum_{i=0}^{n}(-1)^{n-i}
\begin{bmatrix}
n\\ 
i
\end{bmatrix}x^{\color{red}{i}} =\sum_{\color{red}{i=1}}^{n}(-1)^{n-i}
\begin{bmatrix}
n\\ 
i
\end{bmatrix}x^{i}.
\end{eqnarray*}
The zeroth term is zero $\begin{bmatrix}
n\\ 
0
\end{bmatrix}=0$ and can be included or ommited (for convenience.) Have a quick look at the numbers in Wiki
https://en.wikipedia.org/wiki/Stirling_numbers_of_the_first_kind#Table_of_values_for_small_n_and_k
In your formula there is no need for $i \neq j$ and it should look like this 
\begin{eqnarray*}
\prod_{i=1}^{n} (x-i)=\sum_{i=1}^{n+1}(-1)^{n+1-i}
\begin{bmatrix}
n+1 \\ 
i
\end{bmatrix}x^{i-1}.
\end{eqnarray*}
Edit : The Stirling numbers satisfy the following recurrence relation 
\begin{eqnarray*}
\begin{bmatrix} m+1 \\ k \end{bmatrix} = m\begin{bmatrix} m \\ k \end{bmatrix} + \begin{bmatrix} m \\ k-1 \end{bmatrix}. \\
\end{eqnarray*}
If you wish to omit one of the factors, you will need to omit the associated iteration and define a new sequence of numbers
\begin{eqnarray*}
\begin{bmatrix} m+1 \\ k \end{bmatrix} = m\begin{bmatrix} m \\ k \end{bmatrix} + \begin{bmatrix} m \\ k-1 \end{bmatrix} \! m=1,\cdots,j-1 \\
\begin{bmatrix} m+1 \\ k \end{bmatrix} = (m+1)\begin{bmatrix} m \\ k \end{bmatrix} + \begin{bmatrix} m \\ k-1 \end{bmatrix} \! m=j,\cdots,n. \\
\end{eqnarray*}
