Why does $\lim\limits_{n \to \infty } ((1 + x)(1 + {x^2})(1 + {x^4})\ldots(1 + {x^{{2^n}}})) = \frac{1}{{1 - x}}$? Let $\left| x \right| < 1$.       
Why does $$\lim\limits_{n \to \infty } ((1 + x)\cdot(1 + {x^2})\cdot(1 + {x^4})\cdot\ldots\cdot(1 + {x^{{2^n}}})) = \frac{1}{{1 - x}}$$
 A: We have
$$
(1+x)(1+x^2)(1+x^4) \cdots (1+x^{2^n}) = \frac{x^{2^{n+1}}-1}{x-1} \stackrel{n \to \infty}{\longrightarrow} \frac{1}{1-x}
$$
for $n \to \infty$. We used that
$$
x^{2^{n+1}}-1 = (x^{2^n}+1)(x^{2^n}-1) = (x^{2^n}+1)(x^{2^{n-1}}+1)(x^{2^{n-1}}-1) = \cdots = (x^{2^{n}} + 1) (x^{2^{n-1}} + 1) \cdots (x^{4}+1)(x^2+1)(x+1)(x-1),
$$ 
and that $x^{2^{n+1}} \to 0$ for $n \to \infty$.
A: Using $$\displaystyle \lim_{n\rightarrow \infty}\frac{1}{1-x}\left[\overbrace{(1-x)(1+x)}^{(1-x^{2^{1}})}(1+x^2)(1+x^4).......(1+x^{2^{n}})\right]$$
So we get $$\displaystyle \lim_{n\rightarrow \infty}\frac{1}{1-x}\left[\overbrace{(1-x^2)(1+x^2)}^{(1-x^{2^2})}(1+x^4)....(1+x^{2^{n}})\right]$$
In a similar way we get the limit $$\displaystyle = \lim_{n\rightarrow \infty}\frac{1-x^{2^{n+1}}}{1-x}\;,$$ Now Given $-1 <x<1$
So we get limit $$\displaystyle = \frac{1}{1-x}\;,$$ Bcz $x^{2^{n+1}}\rightarrow 0\;,$ When $n\rightarrow \infty$
A: $$(1+x)(1+x^2)(1+x^4)(1+x^8).....=1+x+x^2+x^3+x^4+x^5......$$
then you can use the fact of geometric series
$$\frac{1}{1-x}=1+x+x^2+x^3+x^4+x^5......$$ 
$$|x|<1$$
A: Use $(1 - x)(1 + x)(1 + {x^2}).....(1 + {x^{{2^n}}}) = (1 + {x^{{2^n}}})$
