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.

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
$$\det V= \prod_{1 \leq i < j \leq n}\left(x_i-x_j\right).$$
