Show that the following transformation doesn't change volumes.

$\displaystyle \begin{cases} x_1=u_1 \\ x_2=u_1+u_2 \\ ... \\ x_n=u_1+...+u_n \end{cases}$.

My attempt : I believe I must show that the Jacobian of the transformation is worth $\displaystyle 1$. I'm having a hard time finding the Jacobian though. I've no clue.

$\displaystyle f(x_i)=\sum _{j=1}^{i} u_i$.

$\displaystyle \begin{bmatrix} \frac{\partial f(x_1)}{\partial u_1}, \frac{\partial f(x_2)}{\partial u_1}, ... , \frac{\partial f(x_n)}{\partial u_1} \\ \frac{\partial f(x_1)}{\partial u_2}, \frac{\partial f(x_2)}{\partial u_2}, ... , \frac{\partial f(x_n)}{\partial u_2} \\ ......................... \\ \frac{\partial f(x_1)}{\partial u_n}, \frac{\partial f(x_2)}{\partial u_n}, ... , \frac{\partial f(x_n)}{\partial u_n} \end{bmatrix}$

I know it's wrong, but it's my attempt.