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 $$\lim\limits_{n \to \infty } ((1 + x)\cdot(1 + {x^2})\cdot(1 + {x^4})\cdot\ldots\cdot(1 + {x^{{2^n}}})) = \frac{1}{{1 - x}}$$

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 $n\rightarrow \infty$
