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Algebra Difficulty 4.8 AIME Find the answer

Find (x+1)(x2+1)(x4+1)(x8+1)⋯(x+1)\left(x^{2}+1\right)\left(x^{4}+1\right)\left(x^{8}+1\right) \cdots, where ∣x∣<1|x|<1.

A number or a short expression. Spacing and $ signs are ignored.

Solution

Let S=(x+1)(x2+1)(x4+1)(x8+1)⋯=1+x+x2+x3+⋯S=(x+1)\left(x^{2}+1\right)\left(x^{4}+1\right)\left(x^{8}+1\right) \cdots=1+x+x^{2}+x^{3}+\cdots. Since xS=x+x2+x3+x4+⋯x S=x+x^{2}+x^{3}+x^{4}+\cdots, we have (1−x)S=1(1-x) S=1, so S=11−xS=\frac{1}{1-x}.

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Source: Omni-MATH, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.