An integral inside an integral turns into a product of two integrals? I am struggling to see how after substituting this:
$$\frac{1}{(1+u)^{a+b}}=\frac{1}{\Gamma(a+b)}\int_{0}^{\infty}e^{-(1+u)t}t^{a+b-1}dt$$
into this:
$$B(a,b)=\int_{0}^{\infty}\dfrac{u^{a-1}}{(1+u)^{a+b}}du$$
I will get this:
$$B(a,b)=\frac{1}{\Gamma(a+b)}\int_{0}^{\infty}e^{-t}t^{a+b-1}dt\int_{0}^{\infty}e^{-ut}u^{a-1}du$$
because all I have is this:
$$B(a,b)=\frac{1}{\Gamma(a+b)}\int_{0}^{\infty}u^{a-1}\left[\int_{0}^{\infty}e^{-(1+u)t}t^{a+b-1}dt\right]du$$
Is there a lot of omitted steps here or is there a trick I am failing to catch onto?
The is a part of the derivation on p.25 of this note relating the Beta function to the Gamma function.
Edit:
I give up... I still don't understand but I am starting to seriously question the correctness/necessity of the step mentioned above in the derivation. I think a much more straightforward approach is to multiple left and right side of:
$$B(a,b)=\int_{0}^{\infty}\dfrac{u^{a-1}}{(1-u)^{a+b}}du$$
by the left and right side of:
$${\Gamma(a+b)}=(1+u)^{a+b}\int_{0}^{\infty}e^{-(1+u)t}t^{a+b-1}dt$$
respectively. Then with the change of variable $k=(1+u)t$, arrive at essentially step 2 of the proof linked here.

The hidden step is a change in the order of integration:
$$\int_{0}^{\infty}u^{a-1}\left(\int_{0}^{\infty}e^{-(1+u)t}t^{a+b-1}\,dt\right)du=\int_0^\infty\int_0^\infty u^{a-1}e^{-(1+u)t}t^{a+b-1}\,du\,dt\\
=\int_0^\infty t^{a+b-1}e^{-t}\left(\int_0^\infty u^{a-1}e^{-ut}\,du\right)dt$$
Then, the substitution $u\mapsto u/t$ shows that
$$\int_0^\infty u^{a-1}e^{-ut}\,du=\frac{\Gamma(a)}{t^a} $$
and we're left with
$$\Gamma(a)\int_0^\infty t^{b-1}e^{-t}\,dt=\Gamma(a)\Gamma(b) $$
Edit: I noticed that in the title you mention a product of two integrals. The way the desired result is written could possibly be parsed as a product of integrals but if that were the case, then the equality would be wrong. The correct interpretation should be iterated integration.
