Determinant of Abstract Matrix 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!
 A: Expand the determinant using the usual sum-product formula. Each term in this expansion results by picking some permutation $\pi$. If $\pi(1) = 1$ then the term will be of the form $Cx^{n-1}$, otherwise it will be of the form $Cx^{n-2}$ (since $\pi(1) \neq \pi^{-1}(1)$). Therefore the determinant is of the general form $$ Ax^{n-1} + Bx^{n-2} = x^{n-2} (Ax + B). $$
Substituting $x = 1$, we get that the first two rows are equal, and so the determinant is $0$. So the determinant has the general form $$Cx^{n-2}(x-1).$$
The coefficient of $x^{n-1}$ is clearly equal to the determinant of the $(1,1)$-minor, and if we substitute $x = 1$ we will get a matrix whose determinant is $C$.
As mentioned in another answer, it is very easy to see that the determinant of the matrix is $1$. First we subtract the first row from all other rows. This leaves a lone $1$ on the first row at the first column, so we can erase the first row and column. Continuing this way, we eventually reach the singleton $1$ matrix.
A: I think what you did is almost perfectly correct.  In any case, here is another way that uses induction.  Let $A_n$ refer to the $n\times n$ matrix.  The base case of the $2\times 2$ matrix is $\det A_2 =(x-1)$.
Now, for the inductive step lets look at $A_{n+1}$ and use the linearity of the determinant on the last entry of the last row.  We have that $$\det\left[\begin{array}{ccccc}
1 & 1 & \cdots & 1 & 1\\
1 & x & \cdots & x & x\\
\vdots & \vdots & \ddots & \vdots & \vdots\\
1 & x & \cdots & (n-1)x & (n-1)x\\
1 & x & \cdots & (n-1)x & nx\end{array}\right]=$$ $$\det\left[\begin{array}{ccccc}
1 & 1 & \cdots & 1 & 1\\
1 & x & \cdots & x & x\\
\vdots & \vdots & \ddots & \vdots & \vdots\\
1 & x & \cdots & (n-1)x & (n-1)x\\
1 & x & \cdots & (n-1)x & (n-1)x\end{array}\right]+\det\left[\begin{array}{ccccc}
1 & 1 & \cdots & 1 & 1\\
1 & x & \cdots & x & x\\
\vdots & \vdots & \ddots & \vdots & \vdots\\
1 & x & \cdots & (n-1)x & (n-1)x\\
0 & 0 & \cdots & 0 & x\end{array}\right]$$  The first determinant will be zero since two rows are the same.  The second is $x\det A_n$.
Hope that helps,
A: Let $L$ be the lower triangular matrix whose lower triangular part (including the main diagonal) is filled with ones, and let $e_1=(1,0,0,\ldots,0)^\top$. Then
$$
A=\pmatrix{1&e_1^\top L^\top\\ Le_1&xLL^\top}
=\pmatrix{1\\ &L}
\underbrace{\pmatrix{1&e_1^\top\\ e_1&xI_{n-1}}}_B
\pmatrix{1\\ &L^\top}.
$$
Therefore $\det(A)=\det(B)$. Using Schur complement, if $x$ is an indeterminate, we have $\det(B)=\det(xI_{n-1})\det(1-\frac1x e_1^\top e_1)=x^{n-1}(1-\frac1x)$. Hence $\det(A)=x^{n-2}(x-1)$.
