This is an infinite product which has close similarities to the infinite geometric series. It is well known that if $| x |< 1$ then $$1 + x + x^2 + \dots + x^{n} = \frac{1 - x^{n+1}}{1 - x}$$ and hence (taking limits) we have the sum of the infinite geometric series $$1 + x + x^2 + \dots + x^n + \dots = \frac{1}{1 - x}$$
$$(1 + x)(1 + x^2)(1 + x^4)(1 + x^8)\dots(1 + x^{2^n})\dots = \frac{1}{1 - x}$$ The first step is to evaluate the product of a few terms and then to prove a general result. \eqalign{ (1 - x)(1 + x)(1 + x^2)(1 + x^4)(1 + x^8) &=& (1 - x^2)(1 + x^2)(1 + x^4)(1 + x^8) \\ \; &=& (1 - x^4)(1 + x^4)(1 + x^8) \\ \; &=& (1 - x^8)(1 + x^8) \\ \; &=& (1 - x^{16})} The next step is to use the axiom of mathematical induction to prove the following result: $$(1 + x)(1 + x^2)\dots(1 + x^{2^n}) = \frac{1 - x^{2^{n+1}}}{1 - x}$$ For $n = 1$, \eqalign{ \mbox{LHS} &=& (1 + x)(1 + x^2) \\ \; &=& 1 + x + x^2 + x^3 \\ \mbox{RHS} &=& \frac{1 - x^{2^{1+1}}}{1-x} \\ \; &=& \frac{1 - x^4}{1 - x} \\ \; &=& \frac{(1 - x)(1 + x + x^2 + x^3)}{1 - x} \\ \; &=& 1 + x + x^2 + x^3} So the statement is true for $n = 1$.
Now assume it is true for $n = k$ where $k$ is an integer. $$(1 + x)(1 + x^2)\dots(1 + x^{2^k}) = \frac{1 - x^{2^{k+1}}}{1 - x}$$ It follows that \eqalign{ (1 + x)(1 + x^2)\dots(1 + x^{2^{k+1}}) &=& \frac{1 - x^{2^{k+1}}}{1 - x}(1 + x^{2^{k+1}}) \\ \; &=& \frac{1 - x^{2\times 2^{k+1}}}{1 - x} \\ \; &=& \frac{1 - x^{2^{k+1+1}}}{1 - x}} Hence, by mathematical induction the statement holds for any positive integer value of $n$. Taking limits, where $| x | < 1$ $$\lim_{n\rightarrow\infty}x^{2^n}=0$$ so the formula for the infinite product follows, namely: $$(1 + x)(1 + x^2)(1 + x^4)(1 + x^8)\dots(1 + x^{2^n})\dots = \frac{1}{1 - x}$$