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.

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 $D_n=(x-n)^{n+1}$.
