This is for homework, and I am in need of a hint. Given the product $$ g(z) = \prod_{k=1}^{\infty} \frac{k}{z+k}\left( 1 + \frac{1}{k} \right)^z, $$ I am trying to show that $g(z+1) = zg(z)$. Here is what I have so far.
I tried to get a better sense of $g$, and wrote $$ g(z) = \frac{1}{z+1}\left( 1 + \frac{1}{1} \right)^z\frac{2}{z+2}\left( 1 + \frac{1}{2} \right)^z\frac{3}{z+3}\left( 1 + \frac{1}{3} \right)^z\dotsb. $$ Then, after some manipulation, I found that \begin{align*} g(z+1) &= \frac{1}{(z+1)+1}\left( 1 + \frac{1}{1} \right)^{z+1}\frac{2}{(z+1)+2}\left( 1 + \frac{1}{2} \right)^{z+1}\frac{3}{(z+1)+3}\left( 1 + \frac{1}{3} \right)^{z+1}\dotsb \\ &= \frac{2}{z+2}\left( 1 + \frac{1}{1} \right)^z\frac{3}{z+3}\left( 1 + \frac{1}{2} \right)^z\frac{4}{z+4}\left( 1 + \frac{1}{3} \right)^z\dotsb. \end{align*} Then I tried to calculate $zg(z)$ and see if they match. I found \begin{align*} zg(z) &= \frac{z}{z+1}\left( 1 + \frac{1}{1} \right)^z\frac{2}{z+2}\left( 1 + \frac{1}{2} \right)^z\frac{3}{z+3}\left( 1 + \frac{1}{3} \right)^z\dotsb \\ &= \frac{z}{z+1}\frac{2}{z+2}\left( 1 + \frac{1}{1} \right)^z\frac{3}{z+3}\left( 1 + \frac{1}{2} \right)^z\frac{4}{z+4}\left( 1 + \frac{1}{3} \right)^z\dotsb, \end{align*} where I am doing a rearrangement in the second line. However, now it looks like $g(z+1) = zg(z)$ if and only if $\frac{z}{z+1} = 1$. What am I doing wrong here?
Notice that $g(z+1)$ is the product for $g(z)$ "shifted" by one: that is, $$ g(z+1) = \prod_{k=2}^\infty \frac{k}{z+k}\left(1+\frac{1}{k-1}\right)^z.$$ We can try to rewrite this in terms of the product $\prod\frac{1}{z+k}(1+\frac{1}{k})^z$: $$\prod_{k=2}^\infty \frac{k}{z+k}\left(1+\frac{1}{k-1}\right)^z = \frac{z+1}{1}\prod_{k=1}^\infty \frac{k}{z+k}\left(1+\frac{1}{k}\right)^z$$ where I have multiplied and divided the expression by $\frac{1}{z+1}$ to obtain the right-hand side. Rewriting, $g(z+1) = (z+1)g(z)$.