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 ;)

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 $$\lambda=(x_2-x_3)(x_2-x_4)(x_3-x_4)$$