Assume $d$ positive integer and $\epsilon$ small positive real. How one infers that $$(1-\epsilon)^d\leq e^{-\epsilon d}$$ using the Taylor expansion of exponential function?
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This follows from $1 - \epsilon \le e^{-\epsilon}$, which you can verify using calculus (show $f(x)=e^x - x - 1$ is nonnegative by computing its minimum).
angryavian
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(+1) In fact, one need not use calculus to show that $e^x\ge 1+x$. See THIS ANSWER. – Mark Viola Mar 20 '18 at 16:45