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 ?
 A: 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,
$$
\prod_{k=1}^\infty\left(1+z^k\right)\left(1-z^{2k-1}\right)=1\tag{14}
$$
A: As mr_e_man notes, we have
$$(1+z^n)(1-z^{2n-1})=\frac{(1-z^{2n})(1-z^{2n-1})}{1-z^n}$$
But the numerator covers $1-z^k$ for $k=2n$ and $k=2n-1$, and hence the infinite product is simply one by telescopic cancellation.
This argument relies on the assumption that $\prod_{n=1}^\infty(1-z^n)$ converges, but this can be easily shown by noting that $\sum z^n$ converges.
A: Let the left be $f(z)$. Do associate law: $(1+z)$ vs $(1-z)$, $(1+z^3)$ vs $(1-z^3)$, 
$(1+z^5)$ vs $(1-z^5)$, etc., we see 
$$
f(z)=(1+z^2)(1-z^2)(1+z^4)(1-z^6)(1+z^6)(1-z^{10})...=f(z^2).
$$
Now $f(z)=1+c_nz^n+...$, where $c_n$ is the first nonzero coefficient after $c_0=1$, then $1+c_nz^n+...=f(z)=f(z^2)=1+c_nz^{2n}+...$, thus $c_n=0$. So $f=1$.
A: Hint: Rewrite the product as 
\begin{eqnarray*}
\prod_{k=1}^{\infty} \left( (1-z^{2k-1}) \prod_{j=0}^{\infty} ( 1+z^{(2k-1)2^j}) \right).
\end{eqnarray*}
