Showing Z=XY yields exp(1) distribution I've got the following question, suppose the stochastic vector has a continuous distribution with pdf
$$
f(x,y)=xe^{-x(y+1)},
$$ 
for $x,y>0$. Define $Z=XY$ and show $Z$ has an exponential distrubition with $\beta=1$.
My attempt was defining $g(x,y)=(x \cdot y,y)$, then stating that $g^{-1}=(z/y,y)$, then computing $J(z,y)=1/y$ and hence
$$
f_{Z,Y}=f(g^{-1})\cdot|J(z,y)|=\dfrac{ze^{-(y/z)(y+1)}}{y^2}.
$$ 
When I want to find $f_Z(z)$, I need to integrate it, but I do not find the exp(1) distribution. 
So far, is this the right way? Or should I consider another way?
 A: One way using the characteristic function would be to first take
$$
\begin{align}
\int_0^{\infty}e^{it xy} f(x,y)dy &=xe^{-x}\int_0^{\infty}e^{-(1 - it)xy}dy\\
&= xe^{-x}\frac{1}{(1-it)x} \\
&= e^{-x}\frac{1}{1-it},
\end{align}
$$
and therefore
$$
\mathbb{E}[e^{itXY}] =\int_0^{\infty}\int_0^{\infty}e^{it xy}f(x,y)dydx =\frac{1}{1-it},
$$
which is the characteristic function of an $\mbox{Exp}(1)$ random variable.
As an aside the observation is just that the given joint density factors as
$$
Y|X=x \sim \mbox{Exp}(x), \qquad X \sim \mbox{Exp}(1).
$$
Sorry I haven't been any help with your original attempt, hopefully someone will come along and have a look at that, but this is just an alternative approach that may be of interest.

You're actually really close with your approach, you have a small mistake in that in your notation you should have
$$
f_{Z, Y}(z, y) = \frac{z}{y^2}\exp\left\{-\frac{z}{y}\left(y+1\right)\right\},
$$
note the flip in of the argument of the exponent, then
$$
\begin{align}
ze^{-z}\int_{0}^{\infty}\frac{e^{-zy^{-1}}}{y^2}\operatorname{dy} &= ze^{-z}\int_0^{\infty}e^{-zy^{-1}}\operatorname{d}(y^{-1}) =e^{-z}.
\end{align}
$$
