Hello and Happy holidays. 


I am interested in generalizations of the following product formula for the gamma function 
$\Gamma(z)= \int\_{0}^{\infty} t^{z-1}e^{-t}dt$ when $n \geq 2$:


\begin{align}
\displaystyle\prod\_{k = 1}^{n} \frac{\Gamma(\frac{z}{2^k}+\frac{1}{2})}{\Gamma(\frac{1}{2})} = & \frac{\Gamma(z+1)}{2^{2z(1-\frac{1}{2^n})} \Gamma(\frac{z}{2^n}+\frac{1}{2})} 
\end{align}


Let $H\_1,H\_2,...H\_n \in (0,1)$ and $z \in \mathbb{R^+}$.  

1) Then is it true that the following formula holds for $n \geq 2$?


\begin{align}
\frac{\Gamma(zH\_1 + \frac{1}{2})\Gamma(zH\_1H\_2 + \frac{1}{2}) \dotsb \Gamma(zH\_1H\_2 \dotsb H\_n + \frac{1}{2})}{\prod\_{k=1}^{n} \Gamma(\frac{1}{2})} = 
\end{align}


$\frac{\Gamma(z+1)}{2^{2z(1-H\_1H\_2 \dotsb H\_n)} \Gamma( z H\_1 H\_2 \dotsb H\_n + \frac{1}{2})}$ 


2) As $n$ tends to $\infty$ is the LHS of the last expression finite?


3) Does question 1) hold if $H\_1 = 1$?


(In the context of my research the $H\_i$'s are Hurst parameters from n+1 independent fractional Brownian motions)



Generalizations of a product formula for the gamma function
Certainly not. Let all $H\_i$ tend to zero. By continuity yu get
$$
1=\frac{\Gamma(z+1)}{2^{2z}},
$$
which is false.


