Having seen the answer given by joriki,I think it is the time to share my own answer(I appreciate his answer,and need some time to digest...).
Let \begin{equation} f(x)= \begin{vmatrix} x^{2n+1}&x^{2n}&\cdots&x&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^{2n}&2nx^{2n-1}&\cdots&1&0\\ \vdots&\vdots&\cdots&\vdots&\cdots\\ (2n+1)x_{n}^{2n}&2nx_n^{2n-1}&\cdots&1&0\\ \end{vmatrix} \end{equation}
It is easy to verify that $f(x)$ is a polynomial of degree $4n$.And \begin{equation} f(x_1)=\cdots =f(x_n)=0 \end{equation}
So $(x-x_1)(x-x_2)\cdots (x-x_n)|f(x)$.And
\begin{align*} f'(x)= \begin{vmatrix} x^{2n+1}&x^{2n}&\cdots&x&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(2n+1)x^{2n-1}&(2n-1)2nx^{2n-2}&\cdots&0&0\\ \vdots&\vdots&\cdots&\vdots&\cdots\\ (2n+1)x_{n}^{2n}&2nx_n^{2n-1}&\cdots&1&0\\ \end{vmatrix} \end{align*} So $f'(x_1)=\cdots f'(x_n)=0$.So $(x-x_1)^2(x-x_2)^2\cdots (x-x_n)^2|f(x)$.And
\begin{align*} f''(x)= \begin{vmatrix} (2n+1)x^{2n}&2nx^{2n-1}&\cdots&1&0\\ 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(2n+1)x^{2n-1}&(2n-1)2nx^{2n-2}&\cdots&0&0\\ \vdots&\vdots&\cdots&\vdots&\cdots\\ (2n+1)x_{n}^{2n}&2nx_n^{2n-1}&\cdots&1&0\\ \end{vmatrix}+ \begin{vmatrix} x^{2n+1}&x^{2n}&\cdots&x&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)2n(2n+1)x^{2n-2}&(2n-2)(2n-1)2nx^{2n-3}&\cdots&0&0\\ \vdots&\vdots&\cdots&\vdots&\cdots\\ (2n+1)x_{n}^{2n}&2nx_n^{2n-1}&\cdots&1&0\\ \end{vmatrix} \end{align*}
It is easy to verify that $f''(x_1)=\cdots f''(x_n)=0$.So
\begin{equation} (x-x_1)^3(x-x_2)^3\cdots (x-x_n)^3|f(x) \end{equation} And it is also easy to figure out $f'''(x)$,so it is easy to verify that
\begin{equation} f'''(x_1)=\cdots =f'''(x_n)=0 \end{equation} So \begin{equation} (x-x_1)^4(x-x_2)^4\cdots (x-x_n)^4|f(x) \end{equation}
Because $f(x)$ is a polynomial of degree $4n$,so $f(x)=a(x-x_1)^4(x-x_2)^4\cdots (x-x_n)^4$.According to symmetry ammong $x_0,x_1,\cdots,x_n$ in this determinant,it is easy to verify that
\begin{equation} f(x_0)=c\prod_{n\geq i>j\geq 0}(x_i-x_j)^4 \end{equation} Now we determine the constant $c$.We do it by induction.It is easy to verify that When $n=1$,
\begin{equation} \det \begin{pmatrix} x_0^3&x_0^2&x_0&1\\ x_1^3&x_1^2&x_1&1\\ 3x_0^2&2x_0&1&0\\ 3x_1^2&2x_1&1&0\\ \end{pmatrix}=-(x_0-x_1)^4 \end{equation}Then when $n=2$,Let's see the determinant
\begin{equation} \det\begin{pmatrix} x_0^5&x_0^4&x_0^3&x_0^2&x_0&1\\ x_1^5&x_1^4&*x_1^3&*x_1^2&*x_1&*1\\ x_2^5&x_2^4&*x_2^3&*x_2^2&*x_2&*1\\ 5x_0^4&4x_0^3&3x_0^2&2x_0&1&0\\ 5x_1^4&4x_1^3&*3x_1^2&*2x_1&*1&*0\\ 5x_2^4&4x_2^3&*3x_2^2&*2x_2&*1&*0\\ \end{pmatrix} \end{equation}
The element marked * also form a determinant,we know that the constant of this determinant is -1,so the constant term of the determinant \begin{equation} \det\begin{pmatrix} x_0^5&x_0^4&x_0^3&x_0^2&x_0&1\\ x_1^5&x_1^4&*x_1^3&*x_1^2&*x_1&*1\\ x_2^5&x_2^4&*x_2^3&*x_2^2&*x_2&*1\\ 5x_0^4&4x_0^3&3x_0^2&2x_0&1&0\\ 5x_1^4&4x_1^3&*3x_1^2&*2x_1&*1&*0\\ 5x_2^4&4x_2^3&*3x_2^2&*2x_2&*1&*0\\ \end{pmatrix} \end{equation} is $-1\times (4-5)=1$ ……
So by induction,we know that $c=(-1)^n$.
In my answer I make use of the following result
If \begin{equation} f(x)=\begin{vmatrix} f_{11}(x)&f_{12}(x)&\cdots&f_{1n}(x)\\ f_{21}(x)&f_{22}(x)&\cdots&f_{2n}(x)\\ \vdots&\vdots& &\vdots\\ f_{n1}(x)&f_{n2}(x)&\cdots&f_{nn}(x)\\ \end{vmatrix} \end{equation} Then \begin{equation} f'(x)=\sum_{i=1}^n \begin{vmatrix} f_{11}(x)&f_{12}(x)&\cdots&f_{1n}(x)\\ \vdots&\vdots & &\vdots\\ f'_{i1}(x)&f'_{i2}(x)&\cdots&f'_{in}(x)\\ \vdots&\vdots&&\vdots\\ f_{n1}(x)&f_{n2}(x)&\cdots&f_{nn}(x)\\ \end{vmatrix} \end{equation}