Reducing $\prod \limits_{0 \le j \ne i \le n} \frac{n+1-j}{i-j}$ to $\frac{(n+1)!}{(n+1-i)\cdot i! \cdot (n-i)!}(-1)^{(n-i)}$ How could we show that: $$\prod_{0 \le j \ne i \le n} \frac{n+1-j}{i-j} = \frac{(n+1)!}{(n+1-i)\cdot i! \cdot (n-i)!}(-1)^{(n-i)} .$$
The module suggest we could reduce it by simply writing $$\prod_{0 \le j \ne i \le n} \frac{n+1-j}{i-j} = \frac{(n + 1)n \cdots (n + 1 − (i − 1))(n + 1 − (i + 1)) \cdots 1}{i(i − 1) \cdots 1 · (−1) \cdots (−(n − i))} ,$$ but I am not able to figure out the result from here.
Please explain your answer.
 A: The numerator is
$$\prod_{0 \le j \ne i \le n} (n+1-j) \,.$$ 
By writing it term by term you get 
$$(n+1)n(n-1)....(n+1-(i-1))(n+1-(i+1))...1 = \frac{(n+1)!}{n+1-i} \,.$$
The denominator is 
$$\prod_{0 \le j \ne i \le n} (i-j)= [i(i-1)(i-2)...(i-(i-1))][(i-(i+1))(i-(i+1))...(i-n)] \,.$$
The first bracket is exactly $i!$, while the second is $(-1)(-2)...(-(n-i))$. By taking a minus out of each bracket you get $(-1)^{n-i}(n-i)!$.
Thus the denominator is $(-1)^{n-i}(n-i)!i!$.
Combining those you get exactly the formula Yuqing posted.
A: \begin{equation}
\prod_{0 \le j \ne i \le n} \frac{n+1-j}{i-j} =  \frac{\prod_{0 \le j \ne i \le n}(n+1-j)}{\prod_{0 \le j \ne i \le n}(i-j)}
\end{equation}
we can rewrite this as follows
\begin{equation}
\frac{\prod_{0 \le j \ne i \le n}(n+1-j)}{\prod_{0 \le j \ne i \le n}(i-j)} = \frac{\prod_{0 \le j \le n}(n+1-j)}{(n+1-i) \times \prod_{0 \le j < i}(i-j) \times \prod_{i+1 \le j \le n}-(j-i)}
\end{equation}
You can simplify the above expression to get an expression similar to what you have written. However, there is only $(n+1-i)$ instead of of the $(n+1-i)!$ in the denominator on the right hand side.
A: I assume only $j$ is the running index. Typically the way you have written the product on the left side it means that $i$ and $j$ are both running indices. You could mention this explicitly by using the following notation.
$$\prod_{\substack{j=0\\ j \neq i}}^{n} \frac{n+1-j}{i-j}$$
Split the product into two
$$\prod_{\substack{j=0\\ j \neq i}}^{n} \frac{n+1-j}{i-j} = \prod_{j=0}^{i-1} \frac{n+1-j}{i-j} \times \prod_{j=i+1}^{n} \frac{n+1-j}{i-j}$$
$$\prod_{j=0}^{i-1} \frac{n+1-j}{i-j} = \frac{(n+1)n(n-1)(n-2) \cdots (n-i+2)}{(i)!} = \frac{(n+1)!}{i! (n-i+1)!}$$
$$\prod_{j=i+1}^{n} \frac{n+1-j}{i-j} = (-1)^{n-i} \frac{(n-i)!}{(n-i)!} = (-1)^{n-i}$$
Hence, we get $$(-1)^{n-i} \frac{(n+1)!}{i! (n-i+1)!}$$
