Infinite product equality $\prod_{n=1}^{\infty} \left(1-x^n+x^{2n}\right) = \prod_{n=1}^{\infty} \frac1{1+x^{2n-1}+x^{4n-2}}$ Prove the following equation ($|x|<1$)
$$\prod_{n=1}^{\infty} \left(1-x^n+x^{2n}\right) = \prod_{n=1}^{\infty} \frac1{1+x^{2n-1}+x^{4n-2}}$$
I made this question and I have the following answer but I think it may be incomplete.
If anyone can point out a flaw in my proof or give a better proof then it would be appreciated.
My solution:
$$\begin{align}
&f(x)=\prod_{n=1}^{\infty} \left(1-x^n+x^{2n}\right)\\
&N(p,q)=\{n\in\mathbb N \mid n\ne (2m-1)2^{k-1};m,k\in\mathbb N,2m-1\leq p,k\leq q\}\\
&f(x)(1+x+x^2)=(1+x+x^2)(1-x+x^2)\prod_{n\in N(1,1)}^{\infty} \left(1-x^n+x^{2n}\right)\\
&=(1+x^2+x^4)\prod_{n\in N(1,1)}^{\infty} \left(1-x^n+x^{2n}\right)=(1+x^2+x^4)(1-x^2+x^4)\prod_{n\in N(1,2)}^{\infty} \left(1-x^n+x^{2n}\right)\\
&=(1+x^4+x^8)\prod_{n\in N(1,2)}^{\infty} \left(1-x^n+x^{2n}\right)=(1+x^4+x^8)(1-x^4+x^8)\prod_{n\in N(1,3)}^{\infty} \left(1-x^n+x^{2n}\right)\\
&=(1+x^8+x^{16})\prod_{n\in N(1,3)}^{\infty} \left(1-x^n+x^{2n}\right)=\cdots\\
&=\lim_{k\to\infty}(1+x^{2^k}+x^{2^{k+1}})\prod_{n\in N(1,k)}^{} \left(1-x^n+x^{2n}\right)=\prod_{n\in N(1,\infty)}^{} \left(1-x^n+x^{2n}\right)\\
&\text{Similarly,}\\
&f(x)(1+x+x^2)(1+x^3+x^6)=\prod_{n\in N(3,\infty)}^{} \left(1-x^n+x^{2n}\right)\\
&f(x)(1+x+x^2)(1+x^3+x^6)(1+x^5+x^{10})=\prod_{n\in N(5,\infty)}^{} \left(1-x^n+x^{2n}\right)\\
&\cdots\\
&f(x)\prod_{m=1}^{\infty} \left(1+x^{2m-1}+x^{2(2m-1)}\right)=\lim_{p\to\infty} \prod_{n\in N(p,\infty)}^{} \left(1-x^n+x^{2n}\right)=1
\end{align}$$
(*) Is it obvious that $\{(2m-1)\cdot2^{k-1}\mid m,k\in \mathbb N\}$ is equivalent to $\mathbb N$, or should I also prove it?
Thanks.
 A: We have
\begin{align}
\prod_{n = 1}^\infty (1 - x^n + x^{2n}) &= \prod_{n = 1}^\infty \frac{1+x^{3n}}{1+x^n} \\
&= \prod_{n = 1}^\infty \frac{1 - x^{6n}}{1-x^{3n}}\prod_{n=1}^\infty\frac{1-x^{n}}{1-x^{2n}}\\
& = \prod_{n = 1}^\infty \frac{1}{1-x^{3(2n-1)}}\prod_{n = 1}^\infty (1 - x^{2n-1})\\
&= \prod_{n = 1}^\infty \frac{1-x^{2n-1}}{1-(x^{2n-1})^3}\\
&= \prod_{n = 1}^\infty \frac{1}{1 + x^{2n-1} + x^{4n-2}}.
\end{align}
A: Here is a different/more streamlined/possibly less formal approach. To prove that $$\prod_{n=1}^{\infty}(1-x^n+x^{2n})(1+x^{2n-1}+x^{4n-2})=1$$
Separate the first type of factors into factors coming from even $n$ and odd $n$:
$$\begin{align}
&\prod_{n=1}^{\infty}(1-x^{2n}+x^{4n})(1-x^{2n-1}+x^{4n-2})(1+x^{2n-1}+x^{4n-2})\\
&=\prod_{n=1}^{\infty}(1-x^{2n}+x^{4n})(1+x^{4n-2}+x^{8n-4})
\end{align}$$
Note this is the same as the product from the first line, with $x$ replaced by $x^2$. You can repeat inductively so that the product has higher and higher powers of $x$. If it is convergent at all, it must converge to $1$.
