Find the determinant whose result is $(x-n)^{n+1}$ 
Find the determinant
  $$
\left|\begin{array}{cccccc}{x} & {1} & {} & {} & {} & {} \\ {-n} & {x-2} & {2} & {} & {} & {} \\ {} & {-(n-1)} & {x-4} & {\ddots} & {} & {} \\ {} & {} & {\ddots} & {\ddots} & {n-1} & {} \\ {} & {} & {} & {-2} & {x-2 n+2} & {n} \\ {} & {} & {} & {} & {-1} & {x-2 n}\end{array}\right|
$$

I know the answer is $(x-n)^{n+1}$, and I tried to find it with Gaussian elimination but failed. (That is to reduce the matrix to a matrix with all $(x-n)$ on the diagonal line.)
How to find the determinant? Any method will be appreciated.
 A: Here's a possible method:
From up to down, add the row above to each row:
$$
D_n=\left|\begin{array}{cccccc}{x} & {1} & {} & {} & {} & {} \\ {x-n} & {x-1} & {2} & {} & {} & {} \\ {x-n} & {x-n} & {x-2} & {\ddots} & {} & {} \\ {} & {} & {\ddots} & {\ddots} & {n-1} & {} \\ {} & {} & {} & {x-n} & {x-(n-1)} & {n} \\ {x-n} & {x-n} & {} & {} & {x-n} & {x- n}\end{array}\right|
$$
From left to right, subtract each column from the right column, thus
$$
D_n=\left|\begin{array}{cccccc}{x-1} & {1} & {} & {} & {} & {} \\ {-(n-1)} & {x-1-2} & {2} & {} & {} & {} \\ {} & {-(n-2)} & {x-1-4} & {\ddots} & {} & {} \\ {} & {} & {\ddots} & {\ddots} & {n-1} & {} \\ {} & {} & {} & {-1} & {x-1-2(n-1)} & { } \\ {} & {} & {} & {} & { } & {x- n}\end{array}\right|
=(x-n)D_{n-1}$$
As $D_0=x-n$, we deduce that $D_n=(x-n)^{n+1}$.
A: Here is a different approach using Jordan form.
Setting $x=-\lambda+n$ in the given determinant, the given issue is equivalent to the following one : establish that the characteristic polynomial of:
$$M=\left(\begin{array}{cccccc}{n} & {1} & {} & {} & {} & {} \\ {-n} & {n-2} & {2} & {} & {} & {} \\ {} & {-n+1} & {n-4} & {\ddots} & {} & {} \\ {} & {} & {\ddots} & {\ddots} & {n-1} & {} \\ {} & {} & {} & {-2} & {-n+2} & {n} \\ {} & {} & {} & {} & {-1} & {-n}\end{array}\right)$$
is 

$$(-\lambda)^{n+1}.$$

This will be obtained if we establish that the Jordan decomposition 
$$M=VJV^{-1}\tag{1}$$
gives a matrix $J$ with $J_{ij}=\delta(i-j+1)$ (the only non-zero entries of $J$ are "ones" on the first upper diagonal).
Let us show it, for the sake of simplicity, on the case $n=3$ (the general case being in direct line with this particular case).
$$M=\left(\begin{array}{rrrr}    
3   &  1  &   0 &    0\\
-3   &  1   &  2  &   0\\
0   & -2   & -1  &   3\\
0   &  0  &  -1  &  -3
\end{array}\right), \ J=\left(\begin{array}{cccc}0&1&0&0\\0&0&1&0\\0&0&0&1\\0&0&0&0\end{array}\right), \ V=\left(\begin{array}{rrrr}    
     6  &   6  &   3  &   1\\
   -18  & -12  &  -3  &   0\\
    18  &   6  &   0  &   0\\
    -6   &  0  &   0  &   0
\end{array}\right)$$
Proof : $M$ and $J$ being similar due to relationship (1), they share the same characteristic polynomial which is clearly $(-\lambda)^n.$
Remarks :
1) The general case is not difficult by itself, but is long (and rather uninteresting) to write down. In the general case, as in the particular case, the columns of $V$, starting from the right, are proportional to the successive rows of Pascal's triangle (with alternate signs) . 
2) Matrix $M$ belongs to a category called skew-centronormal matrices with a certain litterature around them.
3) Matrix $M$ is very ill-conditionned.
