Ethan S. answered • 04/18/19

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Not exactly. It is more-so related to the Maclaurin Series for e^{x}. where x=-1. The same Maclaurin Series is useful in proving that the probability of obtaining a derangement is asymptotic. When I try to reason, without proof, why Taylor's Theorem works, I think about the linear approximation of a function (using the first derivative). If the linear approximation is good, then a quadratic approximation would be better, and a cubic approximation even better, so forth and so on. The reason the coefficients are as they are is because think about taking the integral of a power functions, you get x^{n}'s integral is x^{n+1}/(n+1); the integral of this is x^{n+2}/(x+1)(x+2). If you started from a linear term, you would have a denominator of 1*2*3***(n)*(n+1)=(n+1)!. A great visualization of this concept can be found here: https://www.youtube.com/watch?v=3d6DsjIBzJ4.