Showing that $\frac{X_{(i)}}{X_{(n)}},i=1,2,...,n-1$ and $X_{(n)}$ are independent for a population with df $F(y)=y^{\theta}$ 
Let $X_1,X_2,...,X_n$ be i.i.d with df $F(y)=y^{\theta}, 0<y<1, \theta>0$.
Show that $\frac{X_{(i)}}{X_{(n)}}$, for $i=1,2,...,n-1$ and $X_{(n)}$ are independent.

I found the population pdf to be $$f(x)=\theta x^{\theta -1} I_{0<x<1}$$
Now, $$f(x_1,x_2,...x_n)=\theta^n\prod_{i=1}^{n}{x_{i}}^{\theta - 1}$$
Now, the joint distribution of the order statistics:
$$f_{1,2,..,n}(x_1,...x_n)=n! \theta^n\prod_{i=1}^{n}{x_{i}}^{\theta - 1} I_{0<x_1<x_2,...<x_n<1}$$
Now , we take the transformation $Y_i=\frac{X_i}{X_n} $ for $i=1,2,..,n-1$ and $Y_n=X_n$
Have I done correctly so far?
Also, I found out the Jacobian of transformation $={y_n}^{n-1}$.
I am also finding it difficult to get the ranges of $Y_i$'s.
Help!
 A: Joint density of the order statistics $(X_{(1)}=U_1,\cdots,X_{(n)}=U_n)$ is 
$$f_{\mathbf U}(u_1,\cdots,u_n)=n!\,\theta^n\left(\prod_{i=1}^n u_i\right)^{\theta-1}\mathbf1_{0<u_1<\cdots<u_n<1}$$
Transforming $(U_1,\cdots,U_n)\to(Y_1,\cdots,Y_n)$ such that 
$Y_i=\dfrac{U_i}{U_n}$ for $i=1,2,\cdots,n-1$ and $Y_n=U_n$.
The inverse solutions are: $u_n=y_n,u_{n-1}=y_ny_{n-1},u_{n-2}=y_ny_{n-2},\cdots,u_2=y_ny_2,u_1=y_ny_1$
Then, $0<u_1<u_2<\cdots<u_n<1\implies 0<y_1<y_2<\cdots<y_{n-1}<1$ and $0<y_n<1$.
Absolute value of jacobian of transformation is $|J|=y_n^{n-1}$, as you rightly said.
So joint density of $(Y_1,Y_2,\cdots,Y_n)$ is 
$$g(y_1,\cdots,y_n)=n!\,\theta^ny_n^{n\theta-1}\left(\prod_{i=1}^{n-1}y_i\right)^{\theta-1}\mathbf1_{0<y_1<\cdots<y_{n-1}<1,\,0<y_n<1}$$
$$=\left[(n-1)!\theta^{n-1}\left(\prod_{i=1}^{n-1}y_i\right)^{\theta-1}\mathbf1_{0<y_1<\cdots<y_{n-1}<1}\right]\left(n\theta y_n^{n\theta-1}\mathbf1_{0<y_n<1}\right)$$
This implies the independence of $(Y_1,\cdots,Y_{n-1})$ and $Y_n$.
