Identity for Jacobian determinant

Question

For an application in thermodynamics, I want to prove the identity $$ \frac{\partial(u,v)}{\partial(x,y)} = \left[\frac{\partial(x,y)}{\partial(u,v)}\right]^{-1}\, \Leftrightarrow\, \frac{\partial(u,v)}{\partial(x,y)}\cdot \frac{\partial(x,y)}{\partial(u,v)} = 1, $$ where $u = u(x,y)$ and $v = v(x,y)$ are functions of two variables. Thus, we can implicitly solve for $x = x(u,v)$ and $y = y(s,t)$, assuming that all partial derivatives exist and the assumptions of the implicit function theorem are satisfied. Moreover, I denote by $$ \frac{\partial(u,v)}{\partial(x,y)} = \det \begin{pmatrix} \frac{\partial u}{\partial x}|_{y} & \frac{\partial u}{\partial y}|_{x} \\ \frac{\partial v}{\partial x}|_{y} & \frac{\partial v}{\partial y}|_{x} \\ \end{pmatrix}. $$ I want to calculate this directly which gives \begin{align} \frac{\partial(u,v)}{\partial(x,y)}\cdot \frac{\partial(x,y)}{\partial(u,v)} &= \det \begin{pmatrix} \frac{\partial u}{\partial x}|_{y} & \frac{\partial u}{\partial y}|_{x} \\ \frac{\partial v}{\partial x}|_{y} & \frac{\partial v}{\partial y}|_{x} \\ \end{pmatrix}\cdot\det \begin{pmatrix} \frac{\partial x}{\partial u}|_{v} & \frac{\partial x}{\partial v}|_{u} \\ \frac{\partial y}{\partial u}|_{v} & \frac{\partial y}{\partial v}|_{u} \\ \end{pmatrix} \\ &= \frac{\partial u}{\partial x}|_{y}\frac{\partial v}{\partial y}|_{x}\frac{\partial x}{\partial u}|_{v}\frac{\partial y}{\partial v}|_{u}-\frac{\partial u}{\partial x}|_{y}\frac{\partial v}{\partial y}|_{x}\frac{\partial x}{\partial v}|_{u}\frac{\partial y}{\partial u}|_{v}\\ & -\frac{\partial u}{\partial y}|_{x}\frac{\partial v}{\partial x}|_{y}\frac{\partial x}{\partial u}|_{v}\frac{\partial y}{\partial v}|_{u}+\frac{\partial u}{\partial y}|_{x}\frac{\partial v}{\partial x}|_{y}\frac{\partial x}{\partial v}|_{u}\frac{\partial y}{\partial u}|_{v}, \end{align} but I cannot see why the last expression equals $1$.