The original problem is:
Compute the determinant $$\begin{vmatrix} 1^k & 2^k & 3^k & \cdots & n^k \\ 2^k& 3^k & 4^k &\cdots & (n+1)^k \\ 3^k& 4^k & 5^k &\cdots & (n+2)^k \\ \vdots&\vdots &\vdots &\ddots & \vdots\\ n^k& (n+1)^k & (n+2)^k &\cdots &(2n-1)^k \end{vmatrix}$$ where $n\ge 2$ and $0\le k\le n-2.$
The answer is $0$, but I want to find the rank of the above matrix. While solving the original one, I found a beautiful identity. That is, for all $x$: $$\displaystyle \sum _{l=0}^{k+1} {{k+1}\choose l}\cdot (-1)^l\cdot (x+l)^k=0$$
I have two questions:
Base on the above identity, is the rank of the matrix equal to $k+1$?
Is there a name for the above identity?