Determinant of Abstract Matrix
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,
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.