What is the value of the product of Gamma functions \begin{align} \prod_{k=1}^{8} \Gamma\left( \frac{k}{8} \right) \end{align} and can it be shown that the product \begin{align} \prod_{k=1}^{16} \Gamma\left( \frac{k}{8} \right) = \frac{ 3 \Gamma(11) \ \pi}{2^{19}} \zeta(2) \zeta(4) \approx \frac{\pi^{7}}{26} \end{align} is a valid result?
Product of Gamma functions I
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You can use the reflection formula $\Gamma(1-z)\Gamma(z) = \frac{\pi}{\sin(\pi z)}$ in your case for $z\in \{1/8,2/8,3/8\}$, which only leaves the terms $\Gamma(1/2)=\sqrt{\pi}$ and $\Gamma(1)=1$ which are well known, so we have
$$\prod_{k=1}^8 \Gamma\left(\frac{k}{8}\right) = \frac{\pi}{\sin(\pi/8)} \frac{\pi}{\sin(\pi/4)} \frac{\pi}{\sin(3\pi/8)} \sqrt{\pi}$$
I don't know if your second equation is valid, but with $\Gamma(z+1) = z \Gamma(z)$ you can get the left hand side from the initial product, while all the values of $\Gamma$ and $\zeta$ on the right hand side are known, so you can check.
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The multiplication formula says $$\prod_{k=1}^8\operatorname{\Gamma}\left(\frac k8\right)=\prod_{k=0}^7\operatorname{\Gamma}\left(\frac18+\frac k8\right)=(2\pi)^{7/2}\frac{\sqrt8}{8}\operatorname{\Gamma}\left(1\right)=4\pi^{7/2}$$
For the first question, you can simply calculate:
$$\prod_{k=1}^{8}\Gamma\left(\frac{k}{8}\right) = \prod_{k=1}^{8}\frac{8}{k}\cdot\frac{k}{8}! = \frac{8^8}{8!}\cdot\prod_{k=1}^{8}\frac{k}{8}!$$
Or alternatively (even simpler):
$$\prod_{k=1}^{8}\Gamma\left(\frac{k}{8}\right) = \prod_{k=1}^{8}\frac{k-8}{8}!$$