Show $g(z+1) = zg(z)$ 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, $g(z+1) = (z+1)g(z)$.
