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 $|q|<1$ 

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.
