How to find the Jacobian of this transformation

Question

Of this transformation: $$ \begin{aligned} Z_1 &= nX_{(1)} &&\rightarrow &X_{(1)} &= \frac{Z_1}{n} \\ Z_2 &= (n-1)(X_{(2)} - X_{(1)}) &&\rightarrow & X_{(2)} &= \frac{Z_1}{n} + \frac{Z_2}{n-1} \\ &&&\;\;\vdots & \\ Z_n &= (X_{(n)} - X_{(n-1)}) &&\rightarrow &X_{(n)} &= \frac{Z_1}{n} + \frac{Z_2}{n-1}+ \dots + Z_n \\ \end{aligned} $$ Supposing $X_1,\ldots,X_n \sim \operatorname{iid} \text{ Exp}(1)$ and $X_{(1)}, \ldots, X_{(n)}$ are the order statistics. The solution says it is $1/n!$ but I do not really get how if the Jacobian transformation is defined as $$ f(z) = \biggl\lvert \frac{\partial x}{\partial z} \biggr\rvert \, f(x). $$

Answer 1

$$ \begin{bmatrix} \frac{\partial x_{(1)}}{\partial z_1} & \frac{\partial x_{(2)}}{\partial z_1} & ... & \frac{\partial x_{(n)}}{\partial z_1} \\ \frac{\partial x_{(1)}}{\partial z_2} & \frac{\partial x_{(2)}}{\partial z_2} & ... & \frac{\partial x_{(n)}}{\partial z_2} \\ \vdots &\vdots & \ddots & \vdots\\ \frac{\partial x_{(1)}}{\partial z_n} & \frac{\partial x_{(2)}}{\partial z_n} & ... & \frac{\partial x_{(n)}}{\partial z_n} \\ \end{bmatrix} = \begin{bmatrix} \frac1n&\frac1n&...&\frac1n\\ 0&\frac{1}{n-1}&...&\frac{1}{n-1}\\ \vdots & \vdots & \ddots & \vdots \\ 0&0&...&1 \end{bmatrix} $$ $|J|=\frac 1n\cdot \frac{1}{n-1}\cdot ... \cdot 1 = \frac{1}{n!}$