I want to study the asymptotic behaviour of the product
$$
\prod_{i=0}^{n-1} (3i)!.
$$
A first attempt has been to transform the internal factorial to the gamma function, bearing in mind that $\Gamma(n+1)=n!$, and therefore $(3i)!=\Gamma(3i+1)$, for all $i\in\{0,1,\dots, n-1\}$.
However, this brings me to the problem of computing
$$
\prod_{i=0}^{n-1} \Gamma(3i+1),
$$
to which one cannot apply Gauss's Multiplication Formula, which is one of the few ways I have seen to calculate the product of Gammas.
PS: I have already considered approximating the factorial by Stirling's formula, but this is incorrect since this approximation is for "large" values of the factorial, and in this case the value $(3i)!$ may not take "large" values and therefore the asymptotic approximation is not correct.
Best Answer
Beside the Barnes G-function (as mentioned by @Gary in comments), use $$\log (\Gamma (3 i+1))=3 i (\log (i)-1+\log (3))+\frac{1}{2} (\log (i)+\log (2 \pi )+\log (3))+$$ $$\frac{1}{36 i}-\frac{1}{9720 i^3}+O\left(\frac{1}{i^5}\right)$$ Summing from $i=1$ to $i=n-1$, gives for the logarithm of the product $$\frac{1}{2} \left(6 \log (H(n-1))+\log \left(\frac{(1)_{n-1}}{6 \pi }\right)+n \left(n (\log (27)-3)+3+\log \left(\frac{2 \pi }{9}\right)\right)\right)+\frac{H_{n-1}}{36}-\frac{H_{n-1}^{(3)}}{ 9720}$$ where $H(n-1)$ is the hyperfactorial function; the other terms are Pochhammer symbols and generalized harmonic numbers.
This works quite well even for small values of $n$ $$\left( \begin{array}{ccc} n & \text{approximation} & \text{exact} \\ 2 & 1.7917564 & 1.7917595 \\ 3 & 8.3710076 & 8.3710107 \\ 4 & 21.172835 & 21.172838 \\ 5 & 41.160050 & 41.160053 \\ 6 & 69.059321 & 69.059324 \\ 7 & 105.45477 & 105.45477 \\ \end{array} \right)$$
Now, using series expansion for large $n$ $$\color{blue}{\log\left(\prod_{i=0}^{n-1} \Gamma(3i+1)\right)=\frac{3}{4} n^2 \log \left(\frac{9 n^2}{e^3}\right)+n \log \left(\frac{e \sqrt{2 \pi }}{3 n}\right)+}$$ $$\color{blue}{\frac 1{36} \log(n)+\Big[3 \log (A)-\frac{\zeta (3)}{9720}+\frac{\gamma }{36}-\frac{1}{4} \log (18 \pi ) \Big]+}$$ $$\color{blue}{\frac{1}{36 n}+\frac{37}{19440 n^2}-\frac{13}{9720 n^3}+O\left(\frac{1}{n^4}\right)}$$ Repeating the same calculations $$\left( \begin{array}{ccc} n & \text{approximation} & \text{exact} \\ 2 & 1.7917647 & 1.7917595 \\ 3 & 8.3710100 & 8.3710107 \\ 4 & 21.172836 & 21.172838 \\ 5 & 41.160050 & 41.160053 \\ 6 & 69.059321 & 69.059324 \\ 7 & 105.45477 & 105.45477 \\ \end{array} \right)$$
Edit
Using Barnes G-function, given by Mathematica, the asymptotic of the product would be $$\frac{ 2^{\frac{5 n-3}{2}}}{A^3}\times 3^{\frac{(n-1) (3 n-4)}{2} }\times \exp\left(-\frac{9 }{4}n^2+n+\frac{1}{4}+\frac{1}{36 n} \right)\times n^{\frac{3}{2}n^2-n+\frac{1}{36}}\times$$ $$\pi ^{\frac{3 n-1}{2}} \left(G\left(\frac{4}{3}\right) G\left(\frac{5}{3}\right)\right)^{n-2}\times \left(G\left(\frac{7}{3}\right) G\left(\frac{8}{3}\right)\right)^{1-n}$$where $A$ is Glaisher's constant.