Make it Collapse

$\begin{matrix} f\left( x \right) & = & \left( 1+x \right) \left( 1+x^{ 2 } \right) \left( 1+x^{ 4 } \right) \left( 1+x^{ 8 } \right) \dots \\ \left( 1-x \right) f\left( x \right) & = & \left( 1-x \right) \left( 1+x \right) \left( 1+x^{ 2 } \right) \left( 1+x^{ 4 } \right) \left( 1+x^{ 8 } \right) \dots \\ & = & \left( 1-x^{ 2 } \right) \left( 1+x^{ 2 } \right) \left( 1+x^{ 4 } \right) \left( 1+x^{ 8 } \right) \dots \\ & = & \left( 1-{ x }^{ 4 } \right) \left( 1+x^{ 4 } \right) \left( 1+x^{ 8 } \right) \dots \end{matrix}$

If we keep on doing this, every term collapses and we are finally left with

$\left( 1-x \right) f\left( x \right) =1-{ x }^{ { 2 }^{ n } }$

Since there are infinitely many terms, $n$ approaches infinity. Also, since $1>|\frac{1}{3}|$ , $x^{2^{n}}$ approaches 0. Hence we can write the equation again as

$\left( 1-x \right) f\left( x \right) =1 \\ f\left( x \right) = \frac{1}{1-x}$