How to prove the identity $\prod_{n=1}^{\infty} (1-q^{2n-1}) (1+q^{n}) =1 $ for $|q|<1$? Eulers product identity is as follows
\begin{align}
\prod_{n=1}^{\infty} (1-q^{2n-1}) (1+q^{n}) =1
\end{align}
How one can explicitly prove this identity?
Note here $q$ deonotes a complex number satisfying $|q|<1$ 
 A: We have \begin{align}\prod_{n = 1}^\infty (1 - q^{2n-1})(1 + q^n) &= \prod_{n = 1}^\infty \frac{(1 - q^{2n-1})(1 - q^{2n})}{1 - q^n}\\
&= \frac{\prod_{n = 1}^\infty (1 - q^{2n-1})(1 - q^{2n})}{\prod_{n = 1}^\infty (1 - q^n)}\\
&= \frac{\prod_{n = 1}^\infty (1 - q^{2n-1})(1 - q^{2n})}{\prod_{n = 1}^\infty (1 -q^{2n})(1 - q^{2n-1})}\\
&= 1\end{align}
A: For $\lvert q\rvert < 1$, define
$$f(q) = \prod_{n=1}^\infty (1-q^{2n-1})(1+q^n).$$
The product is absolutely and locally uniformly convergent on the open unit disk, hence $f$ is holomorphic there, and the product can be reordered and regrouped as desired. It is clear that $f(0) = 1$, and by reordering and regrouping, we have
$$\begin{aligned}
f(q) &= \prod_{n=1}^\infty (1-q^{2n-1})(1+q^n)\\
&= \prod_{n=1}^\infty(1-q^{2n-1})(1+q^{2n-1})\cdot \prod_{n=1}^\infty (1+q^{2n})\\
&= \prod_{n=1}^\infty \bigl(1-(q^2)^{2n-1}\bigr)\bigl(1+(q^2)^n\bigr)\\
&= f(q^2).
\end{aligned}$$
Since for $0 < \lvert q \rvert < 1$ the set $\{ q^{2^k} : k \in \mathbb{N}\}$ has an accumulation point in the unit disk (namely $0$), it follows that $f$ is constant.
A: Another approach, similar to Daniel's, without an appeal to properties of holomorphic functions.
$$(1-q)(1+q)(1+q^2)\cdots (1+q^{2^n})\cdots = 1$$ 
since the partial product is $1-q^{2^{n+1}}$
.
Substituting $q^{2k-1}$, you get:
$$(1-q^{2k-1})(1+q^{2k-1})(1+q^{(2k-1)\cdot 2})\cdots\left(1+q^{(2k-1)\cdot 2^n}\right)\cdots = 1$$
So $$1=\prod_{k=1}^\infty \left((1-q^{2k-1})\prod_{n=0}^\infty \left(1+q^{(2k-1)2^n}\right)\right)$$
Now, ever positive integer $N$ can be written in exactly one way as $(2k-1)2^n$. So we can rearrange.
