Proving that $\prod_{n=0}^\infty (1 + x^{2^{n}}) = \frac{1}{1-x}$ I would like to verify the truth of my proof that $$\large\prod_{n=0}^\infty (1 + x^{2^{n}}) = \frac{1}{1-x}.$$
Proof. The proof is rather straightforward, just using some observations that two consecutive factors form a sum of a geometric sequence.
\begin{align*}\prod (1 + x^{2^{n}}) &= \prod (1 + x^{2^{2n}})(1+x^{2^{2n+1}}) \\
&= \prod (x^{3\cdot 2^{2n}} + x^{2\cdot 2^{2n}} + x^{1\cdot 2^{2n}} + x^{0\cdot 2^{2n}}) \\
&= \prod \frac{x^{4 \cdot 2^{2n}} - 1}{x^{2^{2n}} - 1}.
\end{align*} Next, we note that $$\large\prod_{n=0}^\infty \frac{x^{4 \cdot 2^{2n}} - 1}{x^{2^{2n}} - 1} = \frac{\prod_{n=0}^\infty x^{4 \cdot 2^{2n}} - 1}{\prod_{n=0}^\infty  x^{2^{2n}} - 1} \qquad (*)$$
Step (*) worries me a little, even though I feel like it is valid. (Would I need to prove that if the limit of the LHS exists, then it's equal to the RHS?)
Then $$\Large \frac{\prod_{n=0}^\infty x^{4 \cdot 2^{2n}} - 1}{\prod_{n=0}^\infty  x^{2^{2n}} - 1} = \frac{\prod_{n=0}^\infty x^{4 \cdot 2^{2n}} - 1}{(1-x)\prod_{n=0}^\infty  x^{2^{2n+2}} - 1} = \frac{1}{1-x},$$ and we are supposedly done.
EDIT: For $|x| < 1$
 A: Hint:
You can make the step $$\prod_n\frac{a_n}{b_n}=\frac{\prod_n a_n}{\prod_n b_n}$$
if and only if the sums
$$A=\sum_n\log a_n,\qquad B=\sum_n\log b_n$$
converge absolutely. Set $a_n=x^{4\cdot2^{2n}}-1$, $b_n=x^{2^{2n}}-1$, and see for which values of $x$ we have absolute convergence of $A, B$.

Edit:
As was pointed out in the comments, we have the definition of an infinite product $\prod_na_n$:
$$\prod_n a_n=\exp\left(\sum_n \log a_n\right),$$
which is equivalent to $$\log\left(\prod_na_n\right)=\sum_n\log a_n.$$
So the product is said to converge iff the sum $\sum_n\log a_n$ converges.
If the sums $A$ and $B$ converge absolutely, then we are justified in saying that
$$A-B=\sum_n\log a_n-\sum_n\log b_n=\sum_n(\log a_n-\log b_n)=\sum_n\log\frac{a_n}{b_n}.\tag 1$$
Since $A=\log \prod_n a_n$ and $B=\log\prod_n b_n$ by definition, we see that
$$A-B=\log \left(\prod_na_n\right)-\log \left(\prod_n b_n\right)=\log\left(\frac{\prod_na_n}{\prod_nb_n}\right).$$
On the other hand, by definition,
$$\sum_n\log\frac{a_n}{b_n}=\log\left(\prod_n\frac{a_n}{b_n}\right).$$
Thus from $(1)$, assuming the products are real-valued,
$$\log\left(\frac{\prod_na_n}{\prod_nb_n}\right)=\log\left(\prod_n\frac{a_n}{b_n}\right)$$
$$\Rightarrow \frac{\prod_na_n}{\prod_nb_n}=\prod_n\frac{a_n}{b_n}.$$
