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 $|q|<1$ 

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.
