Prove an identity including determinant Prove that: 

$$\begin{equation}   \begin{vmatrix}     x_0^{2n+1}&x_0^{2n}&\cdots&x_0&1\\ x_1^{2n+1}&x_1^{2n}&\cdots&x_1&1\\ \vdots&\vdots&\cdots&\vdots\\ x_n^{2n+1}&x_n^{2n}&\cdots&x_n&1\\ (2n+1)x_0^{2n}&2nx_0^{2n-1}&\cdots&1&0\\ \vdots&\vdots&\cdots&\vdots&\cdots\\ (2n+1)x_n^{2n}&2nx_n^{2n-1}&\cdots&1&0\\   \end{vmatrix}=(-1)^{n}\prod_{n\geq i>j\geq 0}(x_i-x_j)^4 \end{equation}$$

where $(x_i^{2n+1})'=(2n+1)x_i^{2n}$,$\cdots$,$x_i'=1$,$1'=0$.
 A: Each column contains the values and first derivatives of a power of $x$ evaluated at $x_0$ to $x_n$. Thus multiplying a column vector of coefficients by this matrix yields a column vector of values and first derivatives of the corresponding polynomial of degree up to $2n+1$ evaluated at $x_0$ to $x_n$. Since this linear evaluation map from the vector space of polynomials of degree up to $2n+1$ to the space of $(2n+2)$-dimensional vectors is an isomorphism if the $x_i$ are distinct (Hermite interpolation establishes a surjection in one direction, and this is a bijection because the dimensions coincide), the product of the matrix with a vector is non-zero unless the vector is zero or two of the $x_i$ coincide; thus the determinant is non-zero unless two of the $x_i$ coincide. Considering the determinant as a polynomial of degree $4n$ in, say, $x_0$, it must be a constant times linear factors $x_0-x_i$ for its roots. By symmetry, each of these factors must appear the same number of times, and thus $4$ times. Applying this argument for each $x_j$ shows that the determinant must be a constant times the product given in the question. To evaluate the constant, consider the term proportional to $x_n^{4n}x_{n-1}^{4(n-1)}\dotso x_{1}^4x_0^0$, which arises from $n+1$ determinants of $2\times2$ matrices containing only one $x_i$ and evaluating to $-x_i^{4i}$; then you just have to check that an odd permutation reorders the matrix such that these $2\times2$ matrices are on the diagonal.
A: There is a sign error in the right-hand side; the term $(-1)^n$ should be $(-1)^{(n+1)(n+2)/2}$.
By the usual formula for the determinant of the Vandermonde matrix,
\begin{equation}   \begin{vmatrix}     x_0^{2n+1}&x_0^{2n}&\cdots&x_0&1\\ x_1^{2n+1}&x_1^{2n}&\cdots&x_1&1\\ \vdots&\vdots&\cdots&\vdots\\ x_n^{2n+1}&x_n^{2n}&\cdots&x_n&1\\ (x_0+h)^{2n+1}&(x_0+h)^{2n}&\cdots&x_0+h&1\\ \vdots&\vdots&\cdots&\vdots&\cdots\\ (x_n+h)^{2n+1}&(x_n+h)^{2n}&\cdots&x_n+h&1\\   \end{vmatrix}=\prod_{i<j} (x_i-x_j)^2 \prod_{i\ne j} (x_i-x_j-h).
\end{equation}
For each of the first $n+1$ rows of the matrix, subtract it from the row $n+1$ rows lower down.  This does not change the determinant of the matrix.  Then divide the right-hand side by $h^{n+1}$, and divide the left-hand side by $h^{n+1}$ by dividing each of the last $n+1$ rows of the matrix by $h$.  Setting $h:=0$ then gives the identity. 
