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How can I calculate without calculator or something like this

the values of $\pi^{e}$ and $e^{\pi}$

in order to compare them ?

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7 Answers 7

up vote 90 down vote accepted

Another proof uses the fact that $\displaystyle \pi \ne e$ and that $e^x > 1 + x$ for $x \ne 0$.

We have $$e^{\pi/e -1} > \pi/e,$$

and so

$$e^{\pi/e} > \pi.$$


$$e^{\pi} > \pi^e.$$

Note: This proof is not specific to $\pi$.

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This is from The Book! Impeccable proof. – Prism Sep 14 '13 at 22:20
@Prism: Thanks! – Aryabhata Sep 17 '13 at 3:27
if x is negative than would your first equation which is e^x > 1 + x be true ? – Murtuza Vadharia Jan 15 at 17:36
@MurtuzaVadharia: Yes, it is true. Consider $f(x) = e^x -1 -x$. It's derivative is $e^x -1$ which is $\lt 0$ for $x \lt 0$ and $\gt 0$ for $x \gt 0$, so $f$ decreases from $-\infty$ to $0$, and increases from $0$ to $\infty$. Since $f(0) = 0$... – Aryabhata Jan 21 at 8:11
I think it's worth mentioning that this works for all positive numbers which aren't $e$. The proof has nothing to do with $\pi$. – NikolajK Feb 7 at 20:48

This is an old chestnut. As a hint, it's easier to consider the more general problem: for which positive $x$ is $e^x>x^e$?

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With the new font it looks like vareps instead of e... Can someone reverse the font?? :/ – AD. Oct 26 '10 at 12:09
What are vareps? – Mariano Suárez-Alvarez Oct 26 '10 at 13:54
\vareps = $\vareps$ ? – J. M. Oct 26 '10 at 14:32
@Mariano, @J.M. I think AD. means \varepsilon = $\varepsilon$ which is an alternate form of $\epsilon$. – Rahul Oct 26 '10 at 16:32
@Rahul Narain: Right, sorry it is my personal abbreviation. – AD. Oct 26 '10 at 20:36

From Proofs without Words.

alt text

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+1 because pictures are always nice to have – J. M. Nov 8 '10 at 15:50
Which software did you use to plot this figure? – whatever Aug 5 '13 at 16:29

Alternatively, we can compare $e^{1/e}$ and $\pi^{1/\pi}$.

Let $f(x) = x^{1/x}$. Then $f'(x) = x^{1/x} (1 - \log(x))/x^2$. Since $\log(x) > 1$ for $x > e$, we see that $f'(x) < 0$ for $e < x < \pi$. We conclude that $\pi^{1/\pi} < e^{1/e}$, and so $\pi^e < e^\pi$.

The same calculation shows that $f(x)$ reaches its maximum at $e^{1/e}$, and so in general $x^e < e^x$.

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Correction: $\text{log}(x)<1$ for $x<e$. – Robert Smith Oct 26 '10 at 15:39
Your argument appears to need $x>e$, but it is easy to also include $x<e$ as well i the $x^e<e^x$ inequality. – Thomas Andrews May 22 '13 at 12:43
It doesn't really need $\log x > 1$. All you need is that the only solution of $\log x = 1$ is $x = e$. – Yuval Filmus May 22 '13 at 14:29

Elaborating Robin's answer take $f(x) = \log{x} - \frac{x}{e}$. We have $$f'(x)= \frac{e-x}{xe}$$ Thus $f'(x)>0$ for $0 < x < e$ and $f'(x) <0$ if $x > e$. Consequently, we have $f(x) < f(e)$ if $x \neq e$.

Exercise: Try to prove this using the same methods: $2^{\sqrt{2}} < e$.

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Prove that the function $f(x)=\frac{e^x}{x^e}, x\geq e$ is strictly increasing on the interval $x\in \left [ e,\pi \right ]$. What is $f(e)$ and $f(\pi)$?

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Let $f (x) =$ $x^\frac1x$ Find value of $x$ such that function gets maximum value For this functions for $x=e$ function will get the maximum value so $e^\frac1e$ is greater than $\pi^\frac1\pi$ so $e^\pi$ is greater than $\pi ^e$.

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please post your answer or any context in formatted version. If you don't know how to format, you may visit this<…; – Subhadeep Dey Aug 12 at 8:15

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