Finding the determinant of a tridiagonal matrix 
$$\begin{vmatrix}x&1&0&0&⋯\\-n&x-2&2&0&⋯\\0&-(n-1)&x-4&3&⋯\\⋮&⋱&⋱&⋱&⋮\\0&⋯&-2&x-2(n-1)&n\\0&0&⋯&-1&x-2n\end{vmatrix}_{(n+1)×(n+1)}$$Find the value of the above determinant.

This problem comes from an advanced algebra book. I want to solve it with elementary transformation knowledge. I have been trying to solve it for a long time.
 A: Edit. Call your matrix $A$ and let $L$ be the zero-indexed Pascal matrix defined by
$$
l_{ij}=\begin{cases}
\binom{n-j}{i}&\text{ when }\ 0\le i\le n-j\le n,\\
0&\text{ otherwise}.
\end{cases}
$$
It is known that $(L^{-1})_{ij}=(-1)^{i+j}l_{ij}$. E.g. when $n=5$,
$$
L=\pmatrix{1&0&0&0&0&0\\ 5&1&0&0&0&0\\ 10&4&1&0&0&0\\ 10&6&3&1&0&0\\ 5&4&3&2&1&0\\ 1&1&1&1&1&1},
\ L^{-1}=\pmatrix{1&0&0&0&0&0\\ -5&1&0&0&0&0\\ 10&-4&1&0&0&0\\ -10&6&-3&1&0&0\\ 5&-4&3&-2&1&0\\ -1&1&-1&1&-1&1}.
$$
One may verify that $A-(x-n)I_{n+1}=L^{-1}NL$ for some nilpotent matrix $N$. More specifically, one may verify that $L(A-(x-n)I_{n+1})=NL$ where
$$
N=\pmatrix{0&1\\ &0&2\\ &&\ddots&\ddots\\ &&&\ddots&n\\ &&&&0}.
$$
In other words, if $V$ is the vector space of polynomials in $y$ of degrees $\le n$ and $D,g,g^{-1}:V\to V$ are the linear operators
\begin{aligned}
D(p)(y)&=p'(y),\\
g(p)(y)&=(1+y)^np\left(\frac{y}{1+y}\right),\\
g^{-1}(p)(y)&=(1-y)^np\left(\frac{y}{1-y}\right),
\end{aligned}
then $A-(x-n)I_{n+1}$ is the matrix representation of the linear map
$f=g^{-1}\circ D\circ g$ with respect to the ordered basis $(1,y,y^2,\ldots,y^n)$. (More explicitly, $f(p)(y)=n(1-y)p(y)+(1-y)^2p'(y)$, but this formula is unimportant here.) Since $D^n=0$, $f^n=g^{-1}\circ D^n\circ g$ is also zero. Hence $f$ and $A-(x-n)I_{n+1}$ are nilpotent and in turn $\det A=(x-n)^{n+1}$.
