special functions - Showing that $Gamma(x)Gamma(y) = Gamma(x+y)beta(x,y)$ via change of variable

From $$\Gamma(x)\Gamma(y)=\int_0^\infty e^{-t}t^{x-1} \left( \int_0^\infty e^{-t} s^{y-1} ds \right) dt,$$ use a change of variable $s=ut$ to show
$$\Gamma(x)\Gamma(y)=\Gamma(x+y)\beta(x,y).$$

Let $s=ut$ so $ds = udt + tdu$ and then

\begin{align*}
\Gamma(x)\Gamma(y) &=\int_0^\infty e^{-t}t^{x-1} \left( \int_0^\infty e^{-s} s^{y-1} ds \right) dt \\
&= \int_0^\infty e^{-t} t^{x-1} \left( \int_0^{\infty} e^{-ut} (ut)^{y-1}(udt+tdu)\right)dt \\
&= \int_0^\infty e^{-t} t^{x-1} \left( \int_0^{\infty} (e^{-ut} u^y t^{y-1})dt+ \int_0^{\infty} (e^{-ut}u^{y-1}t^y)du\right)dt \\
\end{align*}
My next thought was to use integration by parts, but that didn't pan out. Any suggestions?

Note that Gamma Function for $x>0$ we define:

$$\Gamma(x):=\int_0^\infty e^{-t}t^{x-1}dt.$$

And for the Beta Function for $x>0$, $y>0$, we define
$$\beta(x,y):=\int_0^1 t^{x-1}(1-t)^{y-1}dt.$$