Vandermonde determinant for order 4 I'd like to show the case $n=4$ for the Vandermonde-determinant. It should look like this:
$V_4 := \det \begin{pmatrix} 1 & 1 & 1 & 1 \\ x_1 & x_2 & x_3 & x_4 \\ x_1^2 & x_2^2 & x_3^2 & x_4^2 \\ x_1^3 & x_2^3 & x_3^3 & x_4^3 \end{pmatrix} = (x_4 - x_3)(x_4 - x_2)(x_4 - x_1)(x_3 - x_2)(x_3 - x_2)(x_2 - x_1)$.
I tried calculating the $\det(\dots)$ via the first row. However, it ends up in a very large calculation.. I'm sure there's a better way to show especially this case with $n=4$.
Thank you for any help ;)
 A: Let
$$P(x)=\det \begin{pmatrix} 1 & 1 & 1 & 1 \\ x & x_2 & x_3 & x_4 \\ x^2 & x_2^2 & x_3^2 & x_4^2 \\ x^3 & x_2^3 & x_3^3 & x_4^3 \end{pmatrix}$$
then $P$ is a polynomial with degree $3$ and $x_2,x_3,x_4$ are their roots so
$$P(x)=\lambda(x-x_2)(x-x_3)(x-x_4)$$
so the given determinant is
$$P(x_1)=\lambda(x_1-x_2)(x_1-x_3)(x_3-x_4)$$
and finaly to figure out the leading coefficient $\lambda$ we have
$$\lambda=\begin{pmatrix}   1 & 1 & 1 \\   x_2 & x_3 & x_4 \\   x_2^2 & x_3^2 & x_4^2 \end{pmatrix}:=V_3$$
so by simple induction we have
$$\lambda=(x_2-x_3)(x_2-x_4)(x_3-x_4)$$
A: Hint: consider the last row $R_4$, and replace it by $R_4-x_4R_3$, the third row $R_3$ replaced by $R_3-x_4R_2$ and $R_2$ replaced by by $R_2-x_4R_1$. Then after these transformations 
$$V_4=\det\begin{pmatrix}1&1&1&1\\
x_1-x_4&x_2-x_4&x_3-x_4&0\\
x_1(x_1-x_4)&x_2(x_2-x_4)&x_3(x_3-x_4)&0\\
x_1^2(x_1-x_4)&x_2^2(x_2-x_4)&x_3^2(x_3-x_4)&0
\end{pmatrix},$$
which is easier to handle (expand with respect to the last column, factor out $(x_1-x_4)(x_2-x_4)(x_3-x_4)$ and we have to cumpute $V_3$).
A: You can use series of elementary row operations: $-x_1R_3+R_4$, $-x_1R_2+R_3$, $-x_1R_1-R_2$ to get a matrix that is row equivalent to $V_4$. Since it's elementary operations of type 3, the determinant of this matrix is the same as $V_4$, and it is simpler to calculate its determinant. 
