Using the Fourier series expansion for $0 < z < 1$, one has: $$ \ln (\Gamma(z)) = (\tfrac{1}{2}-z)(\gamma+\ln(2)) + (1-z) \ln(\pi)-\frac{1}{2}\ln( \sin(\pi z)) + \frac{1}{\pi} \sum_{n=1}^{\infty} \frac{\sin(2 \pi n z)}{n} \ln(n) .$$
Now taking the general form $$ \prod_{k=1}^{\infty} \frac{(\alpha k-(\alpha-1))^{\frac{1}{\alpha k-(\alpha -1)}}}{(\alpha k-1))^{\frac{1}{\alpha k-1}}} .$$ I have been able to isolate the general form from $\alpha = 2$ to $8$ but I have become stumped upon taking the next power of $2$ and attempting $16$.
For example, $$ \prod_{k=1}^{\infty} \frac{(4k-3)^{\frac{1}{4k-3}}}{(4k-1)^{\frac{1}{4k-1}}} = \left[\frac{ \Gamma(\frac{1}{4}) }{e^{\frac{\gamma}{4}} 2^{\frac{1}{2}} \pi^{\frac{3}{4}} }\right]^{\pi} , \\ \prod_{k=1}^{\infty} \frac{(8k-7)^{\frac{1}{8k-7}}}{(8k-1)^{\frac{1}{8k-1}}} = \left[\frac{\Gamma^{\sqrt{2}}(\frac{1}{8}) \: \Gamma^{1-\frac{1}{\sqrt{2}}} (\frac{1}{4}) \: \sin^{\frac{\sqrt{2}}{2}}(\frac{\pi}{8})}{e^{\frac{\gamma}{4}(1 + \sqrt{2})} \: 2^{\frac{1}{2}(1+\frac{1}{\sqrt{2}})} \: \pi^{\frac{1}{2}(\sqrt{2} + \frac{3}{2} )}}\right]^{\frac{\pi}{2} } .$$
If anyone has any tips or pointers, I would greatly appreciate it.
P.S. Just a heads up; In their product form they have extremely slow convergence.
Let $$f(z)=\pi\left(\ln\Gamma(z)+\left(z-\frac{1}{2}\right)(\gamma+\ln 2)+(z-1)\ln\pi+\frac{1}{2}\ln\sin(\pi z)\right)=\sum_{k=1}^{\infty} \frac{\sin(2 \pi kz)}{k} \ln k$$
Then assuming $\alpha\in\mathbb{N}$ and $\alpha>2$, $$\ln\prod_{k=1}^{\infty} \frac{(\alpha k-\alpha+1)^{\frac{1}{\alpha k-(\alpha -1)}}}{(\alpha k-1)^{\frac{1}{\alpha k-1}}}=\sum_{k=1}^\infty\left(\frac{\ln(\alpha k-\alpha+1)}{\alpha k-\alpha+1}-\frac{\ln(\alpha k-1)}{\alpha k-1}\right)=\sum_{k=1}^\infty s(k)\frac{\ln k}{k}$$ where $$s(k)=\begin{cases} 1 & k\equiv 1\pmod{\alpha}\\ -1 & k\equiv -1\pmod{\alpha}\\ 0 & \text{otherwise} \end{cases}$$ We can express $s(k)$ as a sum of $\sin$ functions using the discrete fourier transform: $$s(k)=\frac{1}{\alpha}\sum_{n=1}^{\alpha-1}s_n\sin\left(\frac{2\pi nk}{\alpha}\right)$$ where $$s_n=\sum_{k=1}^{\alpha-1}s(k)\sin\left(\frac{2\pi nk}{\alpha}\right)=\sin\left(\frac{2\pi n}{\alpha}\right)-\sin\left(\frac{2\pi n(\alpha-1)}{\alpha}\right)=2\sin\left(\frac{2\pi n}{\alpha}\right)$$ Hence, $$\sum_{k=1}^\infty s(k)\frac{\ln k}{k}=\frac{2}{\alpha}\sum_{n=1}^{\alpha-1}\sin\left(\frac{2\pi n}{\alpha}\right)\sum_{k=1}^\infty\sin\left(\frac{2\pi nk}{\alpha}\right)\frac{\ln k}{k}=\frac{2}{\alpha}\sum_{n=1}^{\alpha-1}\sin\left(\frac{2\pi n}{\alpha}\right)f\left(\frac{n}{\alpha}\right)$$ Constant terms in $f$ are annihilated in this formula, and $$\sin(2 x)\ln(\sin(x))=-\sin(2(\pi-x))\ln\sin(\pi-x),$$ so the $\ln\sin$ terms are annihilated as well to give $$\frac{2\pi}{\alpha}\sum_{n=1}^{\alpha-1}\sin\left(\frac{2\pi n}{\alpha}\right)\left(\ln\Gamma\left(\frac{n}{\alpha}\right)+\left(\frac{n}{\alpha}\right)(\gamma+\ln 2\pi)\right)$$ Using $$\sum_{n=1}^{\alpha-1}n\sin\left(\frac{2\pi n}{\alpha}\right)=-\frac{\alpha}{2}\cot\left(\frac{\pi}{\alpha}\right)$$ which comes from taking the imaginary part of $\sum_{n=0}^{\alpha-1} nz^n=\frac{2z}{(1-z)^2} +\frac{(\alpha-1)z}{1-z}$ with $z=\exp\left(\frac{2\pi i}{\alpha}\right)$, we get $$\frac{2\pi}{\alpha}\sum_{n=1}^{\alpha-1}\sin\left(\frac{2\pi n}{\alpha}\right)\left(\frac{n}{\alpha}\right)(\gamma+\ln 2\pi)=-\frac{\pi}{\alpha}(\gamma+\ln 2\pi)\cot\left(\frac{\pi}{\alpha}\right)$$ so we can finally write