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    Power series expansion of the given product of exponential factors

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    By considering appropriate series expansions, prove that the power series expansion of the product of the (infinitely many) exponential factors e^{(x^i)/i}, i = 1, 2, 3, ..., is 1 + x + x^2 ... for |x| <1.

    By expanding each individual exponential factor in the product and multiplying out, also show that the coefficient of x^19 in the power series expansion of the product has the form

    1/(19!) + 1/19 + r/s

    where 19 does not divide s. Deduce that

    18! = -1 (mod 19).

    © BrainMass Inc. brainmass.com October 10, 2019, 8:17 am ad1c9bdddf
    https://brainmass.com/math/number-theory/617268

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    A detailed proof that the power series expansion of the given product of exponential factors is as claimed is provided, and the expansions of the individual exponential factors in the product are used to determine the form of the coefficient of x^{19} in that power series expansion. That, in turn, is used to show that 18! is indeed equivalent to -1 (mod 19).

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