Calculating Vandermonde determinant

You are right. When expanding the determinant with respect to the last column, only the term $$\left| \begin{array}{cccc} 1 & 1 & \cdots & 1\\ x_1 & x_2 & \cdots & x_{n-1} \\ x_1^2 & x_2^2 & \cdots & x_{n-1}^2\\ \cdot & \cdot & \cdots & \cdot \\ x_1^{n-2} & x_2^{n-2} & \cdots & x_{n-1}^{n-2}\\ \end{array} \right|x_n^{n-1} $$

is of degree $n-1$for $x_n$, thus $$\left| \begin{array}{cccc} 1 & 1 & \cdots & 1\\ x_1 & x_2 & \cdots & x_{n-1} \\ x_1^2 & x_2^2 & \cdots & x_{n-1}^2\\ \cdot & \cdot & \cdots & \cdot \\ x_1^{n-2} & x_2^{n-2} & \cdots & x_{n-1}^{n-2}\\ \end{array} \right| = k_n $$