Calculate $\lim_{n\to\infty}[(1+x)(1+x^2)(1+x^4)\cdot\cdot\cdot(1+x^{2n})]$, $|x|<1$

Please help me solving $\lim_{n\to\infty}[(1+x)(1+x^2)(1+x^4)\cdot\cdot\cdot(1+x^{2n})]$, $|x|<1$

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Is the last term supposed to be $(1 + x^{2^n})$? –  Tunococ Sep 10 '12 at 20:34

$\lim_{n\to\infty}[(1+x)(1+x^2)(1+x^4)\cdot\cdot\cdot(1+x^{2n})]$=$\lim_{n\to\infty}\frac{(1+x)(1+x^2)(1+x^4)\cdot\cdot\cdot(1+x^{2n})(1-x)}{1-x}$=$\lim_{n\to\infty}\frac{(1-x^2)(1+x^2)(1+x^4)\cdot\cdot\cdot(1+x^{2n})}{1-x}$=$\lim_{n\to\infty}\frac{(1-x^4)(1+x^4)\cdot\cdot\cdot(1+x^{2n})}{1-x}$=$\cdot\cdot\cdot$=$\lim_{n\to\infty}\frac{(1-x^{2n})(1+x^{2n})}{1-x}$=$\lim_{n\to\infty}\frac{1-x^{4n}}{1-x}$=$\frac{1}{x-1}$$\lim_{n\to\infty}{(1-x^{4n})}=\frac{1}{x-1}$$\cdot 1$=$\frac{1}{x-1}$.
Why do you flip the sign? (And the last term is $1 + x^{2^n}$, right?) –  Tunococ Sep 10 '12 at 20:45
Assuming the last term is $(1 + x^{2^n})$, the product expands out as $$(1 + x)(1 + x^2)(1 + x^4)\ldots(1 + x^{2^n}) = \sum_{k=0}^{2^{n+1} -1} x^k = \frac{1 - x^{2^{n+1}}}{1 - x}.$$
Since $|x| < 1$, this converges to $\frac{1}{1 - x}$ as $n \to \infty$.
The quick and easy way to see this expansion is to note that every number $k$ has a unique binary representation. If the $n$th binary term is $1$, you choose $x^{2^n}$ from the $(1+x^{2^n})$ term, otherwise you choose $1$. –  nayrb Sep 10 '12 at 20:39