Infinite product $\prod_{n=1}^{\infty}(1+z^n)(1-z^{2n-1}) = 1$ I have to show that if $|z| < 1$, $z \in \mathbb{C}$, $$\prod_{n=1}^{\infty}(1+z^n)(1-z^{2n-1}) = 1$$
This exercise if taken from Remmert, Classical Topics in Complex Function Theory.
I don't know where to start. Could anyone give a hint ?

The finite product is
$$
\prod_{k=1}^n\left(1+z^k\right)\left(1-z^{2k-1}\right)
=\prod_{k=n+1}^{2n}\left(1-z^k\right)\tag1
$$
This can be proven by induction. Suppose $(1)$ is true for $n$
$$
\begin{align}
\prod_{k=1}^{n+1}\left(1+z^k\right)\left(1-z^{2k-1}\right)
&=\left(1+z^{n+1}\right)\left(1-z^{2n+1}\right)\prod_{k=1}^n\left(1+z^k\right)\left(1-z^{2k-1}\right)\tag2\\
&=\left(1+z^{n+1}\right)\left(1-z^{2n+1}\right)\prod_{k=n+1}^{2n}\left(1-z^k\right)\tag3\\
&=\left(1-z^{2n+2}\right)\left(1-z^{2n+1}\right)\prod_{k=n+2}^{2n}\left(1-z^k\right)\tag4\\
&=\prod_{k=n+2}^{2n+2}\left(1-z^k\right)\tag5
\end{align}
$$
Explanation:
$(2)$: pull the $k=n+1$ term out front
$(3)$: apply $(1)$ for $n$
$(4)$: pull the $k=n+1$ term out front
$(5)$: bring the $k=2n+1$ and $k=2n+2$ terms inside
Thus, $(1)$ is true for $n+1$. $\large\square$
Since
$$
\begin{align}
\left|\sum_{k=n+1}^{2n}\log\left(1-z^k\right)\right|
&\le\sum_{k=n+1}^{2n}\left|\log\left(1-z^k\right)\right|\tag7\\
&\le\sum_{k=n+1}^\infty\left|\log\left(1-z^k\right)\right|\tag8\\
&\le\sum_{k=n+1}^\infty\frac{|z|^k}{1-|z|^k}\tag9\\
&\le\frac1{1-|z|^{n+1}}\sum_{k=n+1}^\infty|z|^k\tag{10}\\
&=\frac{|z|^{n+1}}{\left(1-|z|\right)\left(1-|z|^{n+1}\right)}\tag{11}
\end{align}
$$
Explanation:
$\phantom{1}(7)$: generalized triangle inequality
$\phantom{1}(8)$: absolute value is non-negative
$\phantom{1}(9)$: $|\log(1+z)|\le\frac{|z|}{1-|z|}$
$(10)$: $\frac1{1-|z|^k}\le\frac1{1-|z|^{n+1}}$
$(11)$: sum of a geometric series 
Equation $(1)$ and inequality $(11)$ say that
$$
\left|\log\left(\prod_{k=1}^n\left(1+z^k\right)\left(1-z^{2k-1}\right)\right)\right|\le\frac{|z|^{n+1}}{\left(1-|z|\right)\left(1-|z|^{n+1}\right)}\tag{12}
$$
Thus,
$$
\log\left(\prod_{k=1}^\infty\left(1+z^k\right)\left(1-z^{2k-1}\right)\right)=0\tag{13}
$$
and therefore,
$$
\prod_{k=1}^\infty\left(1+z^k\right)\left(1-z^{2k-1}\right)=1\tag{14}
$$
