In my book, the Gamma function is defined as $$\Gamma (z)=\frac{1}{g(z)}$$ where $$g(z)=e^{\gamma z}\prod_{k=1}^{\infty}\{(1+z/n)e^{-z/n}\}$$ for a suitable choice of $\gamma$ such that $g(1)=1$. In the proof of the functional equation for this gamma function, what is shown is that $$\frac{g(z)}{zg(z+1)}=e^{-\gamma}\prod_{k=1}^{\infty}\{\frac{n}{n+1}\exp(1/n)\}$$
My question is: Why does this imply that $g(z)=czg(z+1)$ for some constant c?
So the question is:
How do we know $\prod_{n=1}^{\infty}\frac{n}{n+1}e^{\frac{1}{n}}$ converges?
$\prod_{n=1}^\infty a_n $ converges $ \iff \sum_{n=1}^\infty\ln(a_n)$ converges.
So apply the natural log to the product
$$ \begin{align} &\ln\bigg(\prod_{n=1}^{\infty}\frac{n}{n+1}e^{\frac{1}{n}} \bigg) \\&=\sum_{n=1}^\infty \ln(n)-\ln(n+1)+1/n \\ &=\sum_{n=1}^\infty \bigg( \frac{1}{n} -\int_{n}^{n+1} dt/t \bigg) \\& =\lim_{N\to \infty}\sum_{n=1}^N\frac{1}{x}-\int_1^N dt/t \\ & =\gamma \end{align}$$
And this is the famous Euler Mascheroni Constant. You can find a proof that this converges over here.
