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Question as per the title.

Show that $e^x > x^t$ where $x > 0$ and $t < e$

It was suggested that to arrive at the proof, the maclaurin series be used.

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    Please show any working you have done, you're more likely to get positive responses that way.2012-12-11

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Write $x^t= e^{t \log x}$. Then $e^x > x^t$ iff $t < x/\log x$. Now, for $x>0$, the minimum value of $x/\log x$ is $e$, attained at $x=e$.