A closed-form for determinant of matrix

Let an $n\times n$ matrix $A_k$ as shown $$A_k=\left[ \begin{array}{} 1 & x_1 & x_1^2 & \cdots & x_1^{n-2} & x_1^k \\ 1 & x_2 & x_2^2 & \cdots & x_2^{n-2} & x_2^k \\ \vdots & \vdots & \vdots & \cdots & \vdots & \vdots \\ 1 & x_n & x_n^2 & \cdots & x_n^{n-2} & x_n^k \end{array}\right].$$ If $k=n-1$ then the matrix $A_{n-1}$ is called Vandermonde matrix and the determinant of $A_{n-1}$ is $$\det (A_{n-1})= \prod_{1 \mathop \le i \mathop < j \mathop \le n} \left({x_j - x_i}\right).$$

My question is that how to prove: $$\det (A_n)= (\prod_{1\leq i < j \leq n} \left({x_j - x_i}\right))(x_1+x_2+\cdots + x_n)\tag{1}$$

I tried to proof as follows:

We know that if $x_i=x_j$, for some $i\neq j$ then two row of $A_n$ is zero and hence $\prod \left({x_j - x_i}\right)$ is a part of $\det (A_{n-1})$. Moreover, one of the terms of $\det (A_{n-1})$ can be obtained by multiplying of elements of diagonal of $A_n$ which is $x_2x_3^2\cdots x_{n-1}^{n-2}x_{n}^n$, but the degree of $x_n$ in $\prod \left({x_j - x_i}\right)$ is $n-1$ and therefore we need $x_n$. I dont know this argument can be applied to other $x_i$;s or not.