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Algebra Difficulty 2.2 Junior Find the answer

If s=(1+2−132)(1+2−116)(1+2−18)(1+2−14)(1+2−12)s=(1+2^{-\frac{1}{32}})(1+2^{-\frac{1}{16}})(1+2^{-\frac{1}{8}})(1+2^{-\frac{1}{4}})(1+2^{-\frac{1}{2}}), then ss is equal to
(A) 12(1−2−132)−1\textbf{(A) }\textstyle{\frac{1}{2}}(1-2^{-\frac{1}{32}})^{-1}(B) (1−2−132)−1\textbf{(B) }(1-2^{-\frac{1}{32}})^{-1}(C) 1−2−132\textbf{(C) }1-2^{-\frac{1}{32}}(D) 12(1−2−132)\\ \textbf{(D) }\textstyle{\frac{1}{2}}(1-2^{-\frac{1}{32}})(E) 12\textbf{(E) }\frac{1}{2}

Solution

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.