Show the Jacobian back to itself is 1 I'm working through problems for revision but I've lost the solutions.

Show that $\frac{\partial (x, y)}{\partial (u, v)} \frac{\partial (u,
v)}{\partial (x, y)} = 1$.

My attempt at a solution: Intuitively this makes sense because we convert coordinates from $(x,y) \rightarrow (u,v) \rightarrow (x,y)$ so we should end up with a scale factor of 1. I showed in a previous part of this problem that $$\frac{\partial (r, s)}{\partial (u, v)} \frac{\partial (u, v)}{\partial (x, y)} = \frac{\partial (r, s)}{\partial (x, y)} $$ so the original equation becomes $$\frac{\partial (x, y)}{\partial (u, v)} \frac{\partial (u,v)}{\partial (x, y)} = \frac{\partial (x, y)}{\partial (x, y)} = \frac{\partial x}{ \partial x}\frac{\partial y}{ \partial y} - \frac{\partial y}{ \partial x} \frac{\partial x}{ \partial y} = 1 \cdot 1 - 1 = 0.  $$
I'm not sure what I'm missing here. 
 A: This result follows directly from the chain rule and a basic property of determinants. I'll give details for the 2-variable case that you're working with.
$$\dfrac{\partial(x,y)}{\partial(u,v)} = \mathrm{det}\begin{pmatrix} x_u & y_u \\ x_v & y_v \end{pmatrix} = x_uy_v-y_ux_v$$
Likewise, $\dfrac{\partial(u,v)}{\partial(x,y)} = u_xv_y-v_xu_y$. 
Therefore, $\dfrac{\partial(x,y)}{\partial(u,v)} \cdot \dfrac{\partial(u,v)}{\partial(x,y)} = (x_uy_v-y_ux_v)\cdot (u_xv_y-v_xu_y)$ 
$ = x_uy_vu_xv_y-y_ux_vu_xv_y-x_uy_vv_xu_y+y_ux_vv_xu_y$
$ = x_uu_xy_vv_y-x_vv_yy_uu_x-x_uu_yy_vv_x+x_vv_xy_uu_y+(x_uu_xy_uu_y-x_uu_yy_uu_x)+(x_vv_xy_vv_y-x_vv_yy_vv_x)$
$ = (x_uu_x+x_vv_x)(y_uu_y+y_vv_y)-(x_uu_y+x_vv_y)(y_uu_x+y_vv_x)$
$=\dfrac{\partial x}{\partial x}\dfrac{\partial y}{\partial y} - \dfrac{\partial x}{\partial y}\dfrac{\partial y}{\partial x} = 1\cdot 1-0 \cdot 0 = 1$
where the next to last equality follows from several applications of the chain rule.
Here's a quick version (which works for transformations of $n$ variables not just $n=2$)...
Suppose $J_T$ is the Jacobian matrix of some coordinate transform $T$. It is easy to see that $J_I=I$ (the Jacobian matrix of the identity transformation is the identity matrix). 
The chain rule says that $J_{S\circ T} = J_SJ_T$ (where $S$ and $T$ are maps).
Notice $I=J_I=J_{T \circ T^{-1}}=J_TJ_{T^{-1}}$. Taking determinants we get $1=\mathrm{det}(I)=\mathrm{det}(J_TJ_{T^{-1}})=\mathrm{det}(J_T)\mathrm{det}(J_{T^{-1}})$. 
Thus $\mathrm{det}(J_{T^{-1}})=\dfrac{1}{\mathrm{det}(J_T)}$.
In particular, $\dfrac{\partial(x,y)}{\partial(u,v)} \cdot \dfrac{\partial(u,v)}{\partial(x,y)} = \dfrac{\partial(x,y)}{\partial(u,v)} \cdot \dfrac{1}{\dfrac{\partial(x,y)}{\partial(u,v)}} = 1$.
So for example, the Jacobian (determinant) of the transform from rectangular to polar coordinates is $r$. So the Jaocbian for the transform from polar back to rectangular coordinates is $\dfrac{1}{r} = \dfrac{1}{\sqrt{x^2+y^2}}$.
