Determinant of matrix $(x_j^{n-i}- x_j^{2n-i})_{i,j=1}^{n}$ Good evening all, I am determined to determine this determinant:
$$D = \det{\left[x_j^{n-i} - x_j^{2n-i}\right]_{i,j=1}^{n}}$$
Looking at the smaller cases, leads me to believe that
$$D = \prod_{1 \leq i < j \leq n}\left(x_i-x_j\right)\prod_{i=1}^n \left(1-{x_i}^n\right)$$
although I am having trouble showing this. I know that, since the determinant is an alternating function in the variables $x_1,\dots x_n$ it follows that
$$ \frac{D}{\displaystyle\prod_{1 \leq i < j \leq n}\left(x_i-x_j\right)} $$
is a symmetric polynomial of degree $n^2$ (the degree of D minus the degree of the Vandermonde part).
How can I show that this symmetric polynomial is exactly $\prod_{i=1}^n \left(1-{x_i}^n\right)$ ?
Your help is, as always, much appreciated.
 A: We would like to know the determinant
$ \det M$ of the matrix $n\times n$-matrix
$$M_{ij} = x_j^{n-i} - x_j^{2n-i}.$$
Note that $$M_{ij} = (1- x_j^n) x_j^{n-i} = (1-x_j^n) V_{ij}.$$ It is easy to see (column operation) that 
$$ \det M = \det V \,\prod_{j=1}^n (1-x_j^n).$$
Now $V_{ij} = x^{n-i}_j$ is a Vandermonde matrix whose determinant is known (and can be proven by induction) to be
$$\det V= \prod_{1 \leq i < j \leq n}\left(x_i-x_j\right).$$
A: The $(i,j)$-entry can be expressed as: $x_j^{n-i}-x_j^{2n-i}=x_j^{n-i}(1-x_j^n)$. 
So column $j$ has the common factor $(1-x_j^n)$. Using multilinearity of the determinant pull these factors out (from each column). Then you have
$$ D = \prod_{j=1}^n (1-x_j^n) \; \mathrm{det} \begin{bmatrix} x_1^{n-1} & x_2^{n-1} & \cdots & x_n^{n-1} \\ x_1^{n-2} & x_2^{n-2} & \cdots & x_n^{n-2} \\ \vdots & \vdots & & \vdots \\ x_1 & x_2 & \cdots & x_n \\ 1 & 1 & \cdots & 1 \end{bmatrix} $$
This an upside down Vandermonde. Interchanging rows $1$ and $n$, $2$ and $n-1$, etc. gives us the sign $(-1)^{\ell}$ where $\ell$ is the floor of $n/2$. Then using the Vandermonde determinant formula finishes the job. 
So I think your result is off by sign.
As Fabian pointed out, the OP's product is $x_i-x_j$ for $i<j$ instead of $x_j-x_i$ (that appear in the standard Vandermonde formula), so the sign cancels back out and the result follows (no adjustment needed).
