Product of Gamma functions II What is the value of the product of Gamma functions
\begin{align}
\prod_{k=1}^{10} \Gamma\left(\frac{k}{10}\right)
\end{align}
and can it be shown that
\begin{align}
\prod_{k=1}^{20} \Gamma\left(\frac{k}{10}\right) \approx \frac{\pi^{9}}{54}
\end{align}
and
\begin{align}
\prod_{k=1}^{40} \Gamma\left(\frac{k}{10}\right) \approx \left( 6 + \frac{625}{4501}\right) \pi^{18}.
\end{align}
 A: Since @robjohn has done an excellent job at providing an answer in detail I will add my comments as an additional solution. The results presented here follow the numbering in robjohn's work.
In robjohn's equation (12) the factor 
\begin{align}
\frac{9! \ \pi^{9}}{2 \cdot 5^{10}}
\end{align}
has been provided. Numerically this can be seen as $.018579456\cdots \ \pi^{9}$. By comparing this to that of $1/54 = .0185185\cdots$ one can make a good approximation by
\begin{align}
\prod_{k=1}^{20} \Gamma\left(\frac{k}{10}\right) = \frac{9! \ \pi^{9}}{2 \cdot 5^{10}} \approx \frac{\pi^{9}}{54}. \tag{14}
\end{align}
Another possible value is $20/1077 = .0185701021\cdots$ which leads to the statement
\begin{align}
\prod_{k=1}^{20} \Gamma\left(\frac{k}{10}\right) = \frac{9! \ \pi^{9}}{2 \cdot 5^{10}} \approx \frac{20 \ \pi^{9}}{1077}. \tag{15}
\end{align}
From equation (13) the factor is
\begin{align}
\frac{2^{18} \ (10)!(20)!(30)!}{10^{62}} = 6.138858832\cdots = 6 + .138858832\cdots . 
\end{align}
Since $625/4501 = .1388580315\cdots$ then to a fair approximation it can be stated
\begin{align}
\prod_{k=1}^{40} \Gamma\left(\frac{k}{10}\right) \approx \left( 6 + \frac{625}{4501} \right) \pi^{18}. \tag{16}
\end{align}
