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

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 $n \to \infty$.
