Let

$A\_n = \begin{pmatrix}1 & x\_1 & x\_1^2 & \dots &{x\_1}^{n-1} \\1 & x\_2 & x\_2^2 & \dots & {x\_2}^{n-1} \\\vdots & \vdots & \vdots & & \vdots \\1 & x\_n & x\_n^2 & \dots &{x\_n}^{n-1} \\\end{pmatrix}$
We know that $det(A\_n)=\prod\_{1 \leq j < i \leq n} (x\_i-x\_j)$.

In case of $n=3$

$A\_3 = \begin{pmatrix}1 & x\_1 &{x\_1}^{2} \\1 & x\_2 &{x\_2}^{2} \\ 1 & x\_3 &{x\_3}^{2} \\\end{pmatrix}$

However I get something entirely different when applying Sarrus. With Sarrus I get $x\_2\*{x\_3}^2+x\_1\*{x\_2}^{2}+{x\_1}^{2}\*{x\_3}-{x\_1}^2\*x\_2-x\_1\*{x\_3}^2-{x\_2}^{2}\*x\_3$

when applying the given formula $det(A\_3)=\prod\_{1 \leq j < i \leq 3} (x\_i-x\_j)=(x\_3-x\_2)(x\_3-x\_1)(x\_2-x\_1)$

I do not know whether I applied the product with the indices correctly. Could someone help me?