Given an $n \times n$ matrix $A$, where $x$ is any real number:
$A = \left[ \begin{array}{ c c c c c c c c } 1 & 1 & 1 & 1 & 1 & 1 & \cdots & 1 \\ 1 & x & x & x & x & x & \cdots & x \\ 1 & x & 2x & 2x & 2x & 2x & \cdots & 2x \\ 1 & x & 2x & 3x & 3x & 3x & \cdots & 3x \\ 1 & x & 2x & 3x & 4x & 4x & \cdots & 4x \\ 1 & x & 2x & 3x & 4x & 5x & \cdots & 5x \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & x & 2x & 3x & 4x & 5x & \cdots & (n-1)x \end{array} \right]$
Find the determinant.
By using $n=2,3,4,5,...$ and random $x=1,2,3,4...$ I have found that $det(A) = (x-1)(x)^{n-2}$ through observing a pattern.
However, I would like to be able to prove this through a proof, yet I have no idea where to start.
When I try to solve for the determinant using the abstract matrix A and using the property that the determinant of a square matrix is $(-1)^r * (\text{products of pivots})$, where r is the number of row interchanges, my answer is of the form $(x-1)(x)(x)(x)(x)...(n-?)x$ where "?" depends on how many rows I include in the abstract form of A. How do I show that $(x)(x)(x)...(n-?)x$ equals $(x)^{n-2}$?
Here is my work: http://i.imgur.com/tinDw.jpg
Any hints? Thanks for the help!