## Which is bigger: π^e or e^π?

The question of whether $\pi^e$ or $e^{\pi}$ is larger is a classic one.  Of course, with a calculator it is easy to see what the answer is.  But how would you answer the question without a calculator?

There are lots of interesting ways to tackle the question.  For this (short) post I would like to highlight one I saw on math.SE, given by Aryabhata.  We need the assumptions that $\pi > e$ and $e^x > x+1$ for $x > 0$.  (One way to view the latter claim is that, for $x > 0$, $e^x$ is larger than the first two terms of its Maclaurin series.  Since all the rest of the terms of the Maclaurin series are positive, and the Maclaurin series converges to $e^x$, we get that $e^x > x+1$ for $x > 0$.)

On to the proof.  Since $\pi > e$, $\pi/e - 1 > 0$.  Applying the second assumption, then, we get $\displaystyle e^{\pi/e-1} > \pi/e$.  This means that $\displaystyle e^{\pi/e} > \pi$, which in turn implies that $\displaystyle e^{\pi} > \pi^e$.

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