Proving the relationship between Gamma and Beta functions without Fubini theorem is it possible to prove the identity
$$\forall x > 0,\ \forall y > 0,\quad B(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$
where $\Gamma$ is the Gamma function and $B$ is the Beta function without using the Fubini theorem (the standard proof using it and polar coordinates)?
It am looking for an elementary proof of this identity...
 A: We can prove the identity in a different way using Euler's infinite product representation of the gamma function,
$$\Gamma(x) = \lim_{n \to \infty}\Gamma_n(x) = \lim_{n \to \infty} \frac{n!n^x}{(x+1)(x+2) \cdots (x+n)}$$ 
The beta function defined as $\displaystyle B(x,y) = \int_0^1 t^{x-1}(1-t)^{y-1} \, dt$ satisfies the recurrence relation 
$$ \tag{1}B(x,y) = \frac{x+y}{y}B(x,y+1)$$
The proof of (1) involves integration by parts and can be furnished if desired. Applying (1) $n$ times we get
$$\tag{2}B(x,y) = \frac{(x+y)(x+y+1) \cdots (x + y + n)}{y(y+1) \cdots (y +n)}B(x,y+n+1)  \\ =\frac{\Gamma_n(y)}{\Gamma_n(x+y)}n^{x}B(x,y+n+1)$$
Note that
$$n^{x}B(x,y+n+1) = n^{x}\int_0^1t^{x-1}(1-t)^{y+n} \, dt,$$
and after changing variables with $s = nt$ we get
$$\tag{3}n^{x}B(x,y+n+1) = \int_0^ns^{x-1}\left(1-\frac{s}{n}\right)^{y+n} \, ds = \int_0^ns^{x-1}\left(1-\frac{s}{n}\right)^{n} \left(1-\frac{s}{n}\right)^{y}\, ds $$
Substituting into (2) using (3) yields
$$\tag{4}B(x,y) = \frac{\Gamma_n(y)}{\Gamma_n(x+y)}\int_0^ns^{x-1}\left(1-\frac{s}{n}\right)^{n} \left(1-\frac{s}{n}\right)^{y}\, ds$$
By the dominated convergence theorem,
$$\lim_{n \to \infty}\int_0^ns^{x-1}\left(1-\frac{s}{n}\right)^{n} \left(1-\frac{s}{n}\right)^{y}\, ds  =  \lim_{n \to \infty}\int_0^{\infty}s^{x-1}\left(1-\frac{s}{n}\right)^{n} \left(1-\frac{s}{n}\right)^{y}1_{[0,n]}\, ds  \\ = \int_0^\infty s^{x-1}e^{-s}\, ds ,$$
and, since $\lim_{n \to \infty} \Gamma_n(z) = \Gamma(z)$, upon taking the limit of both sides of (4) as $n \to \infty$ we obtain
$$B(x,y) = \frac{\Gamma(y)}{\Gamma(x+y)}\int_0^\infty s^{x-1}e^{-s}\, ds = \frac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)}$$
