proving: $(1+x)(1+x^2)(1+x^3)\ldots = \frac{1}{(1-x)(1-x^3)(1-x^5)\ldots}$ I need to prove:
$(1+x)(1+x^2)(1+x^3)\ldots = \frac{1}{(1-x)(1-x^3)(1-x^5)\ldots}$  for  $0<x<1$
First i proved: $(1)$ $\prod_{n=0}^{\infty}(1+x^{2^n}) = \frac{1}{1-x}$ for $ |x|<1$
Now my attemp was:
$(1+x)(1+x^2)(1+x^3)\ldots = \prod_{k \in \mathbb{N}_{odd}}\prod_{n=0}^{\infty}(1+(x^k)^{2^n})= \frac{1}{(1-x)(1-x^3)(1-x^5)\ldots}$
where the last equality derives form (1) by substituting $x=x^k$ meaning: $\prod_{n=0}^{\infty}(1+(x^k)^{2^n}) = \frac{1}{1-x^k}$
I wanted to make sure this is a formal enough solution, this is our first time using infinite product and i am not sure if this is correct.
 A: Here is a way to improve your answer.  First, you should show that the product
$$f(x):=(1+x)(1+x^2)(1+x^3)\ldots$$
converges absolutely for complex numbers $x$ such that $|x|<1$.  Then, you can rearrange the product to get the final result as you did.
Now, how do we show absolute convergence of $f(x)$?  You can try to prove that
$$\ln\big(f(x)\big)=\sum_{k=1}^\infty\,\ln(1+x^k)$$
converges absolutely.  From
$$\ln(1+t)=t+\mathcal{O}(t^3)$$
for $t\in\mathbb{C}$ with $|t|<1$, we have
$$\sum_{k=1}^\infty\,\big|\ln(1+x^k)\big|=\sum_{k=1}^\infty\,\Big(|x|^k+\mathcal{O}\big(|x|^{3k}\big)\Big)=\frac{|x|}{1-|x|}+\mathcal{O}\left(\frac{|x|^3}{1-|x|^3}\right)\,.$$
Thus, $\ln\big(f(x)\big)$ converges absolutely.
Here is an explanation why it is important to check whether your product converges absolutely before rearranging it.  Let us take a look at this example:
$$P:=\prod_{n=1}^\infty\,\left(1+\frac{(-1)^n}{1+\left\lceil\frac{n}{2}\right\rceil}\right)=\left(1-\frac{1}{2}\right)\left(1+\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1+\frac13\right)\cdots\,.$$
If you define the $m$-th partial product to be
$$P_m^1:=\underbrace{\left(1-\frac12\right)\left(1+\frac12\right)\left(1-\frac13\right)\left(1+\frac13\right)\left(1-\frac14\right)\left(1+\frac14\right)\cdots}_{m\text{ pairs of parentheses}}\,,$$
then
$$P_{2m}^1=\frac{(m+2)}{2(m+1)}\,,$$
suggesting that
$$\lim_{m\to\infty}\,P_{m}^1=\frac12\,.$$
If you define the $m$-th partial product to be
$$P_m^2:=\underbrace{\left(1-\frac12\right)\left(1-\frac13\right)\left(1+\frac12\right)\left(1-\frac14\right)\left(1-\frac15\right)\left(1+\frac13\right)\cdots}_{m\text{ pairs of parentheses}}\,,$$
then
$$P_{3m}^2=\frac{(m+2)}{2(2m+1)}\,,$$
suggesting that
$$\lim_{m\to\infty}\,P_{m}^2=\frac14\,.$$
If you define the $m$-th partial product to be
$$P_m^3:=\underbrace{\left(1-\frac12\right)\left(1+\frac12\right)\left(1+\frac13\right)\left(1-\frac13\right)\left(1+\frac14\right)\left(1+\frac15\right)\cdots}_{m\text{ pairs of parentheses}}\,,$$
then
$$P_{3m}^3=\frac{(2m+1)}{2(m+1)}\,,$$
suggesting that
$$\lim_{m\to\infty}\,P_{m}^3=1\,.$$
