Reading the answer here, it seems like the relation:
$$B(t,s)=\frac{\Gamma(t)\Gamma(s)}{\Gamma(t+s)} \tag{1}$$
follows from the semi-group property of the Gamma distribution since "the integral of a convolution of two functions is the product of their integrals".
The semi-group property being spoke of here:
$$\gamma_t*\gamma_s=\gamma_{t+s} \tag{2}$$
where:
$$\gamma_t(x):=\frac{x_+^{t-1}e^{-x}}{\Gamma(t)}$$
Seems to indicate there is some elegant proof for equation (1) using equation (2). Is this true or am I imagining things? And if there is, can someone show me what it might be?
From Wikipedia:
With $f(u) = e^{-u} u^{x-1} 1_{u > 0}$ and $g(u) = e^{-u} u^{y-1} 1_{u > 0}$, we have $$\Gamma(x) \Gamma(y) = \int_{-\infty}^\infty f(u) \, du \cdot \int_{-\infty}^\infty g(u) \, du = \int_{-\infty}^\infty (f * g)(u) \, du = B(x,y) \Gamma(x+y).$$
To show the last equality, note that $$(f*g)(u) = \int_{-\infty}^\infty f(t) g(u-t) \, dt = \int_0^u e^{-u} t^{x-1} (u-t)^{y-1} \, dt = e^{-u} u^{x+y-1} \int_0^1 s^{x-1} (1-s)^{y-1} \, ds$$ where the last step comes from the substitution $t=us$.