I looked at a couple of ways of examining this using series expansions. A difficulty that I noted in a comment to **mathlove**'s answer is that this requires rather careful selection of estimators, since the permissible "error budget" is *very* limited if we are to show that the difference is less than 1 , when its actual value is about 0.682. Another issue is that we would like to use rational approximations in a way that makes these computations look credibly like ones that can be managed without a calculator. I use some "number facts" here that are either generally familiar or can be found readily enough (while these can be obtained using a calculator, that can be avoided if one is acquainted with approximate values); it proves to be very important to use values that closely approximate the needed numbers.

We can attempt to estimate an upper limit the difference $ \ e^{\pi} \ - \ \pi^e \ $ by setting up an inquality in which $ \ e^{\pi} \ $ is overestimated and $ \ \pi^e \ $ is underestimated. In this first one, we will start from

$$ \left( \frac{68}{25} \right)^{22/7} \ > \ e^{\pi} \ > \ \pi^{e} \ > \ \left( \frac{157}{50} \right)^{19/7} \ \ $$

$$ \Rightarrow \ \ \left( 3 \ - \ \frac{7}{25} \right)^{3 + \frac{1}{7}} \ > \ e^{\pi} \ > \ \pi^{e} \ > \ \left( 3 \ + \ \frac{7}{50} \right)^{3 - \frac{2}{7}} \ \ $$

$$ \Rightarrow \ \ 3^3 \ \cdot \ 3^{1/7} \ \cdot \left( 1 \ - \ \frac{7}{75} \right)^3 \ \left( 1 \ - \ \frac{7}{75} \right)^{\frac{1}{7}} \ > \ e^{\pi} \ > \ \pi^{e} \quad > \ 3^3 \ \cdot \ 3^{-2/7} \ \cdot \left( 1 \ + \ \frac{7}{150} \right)^3 \ \left( 1 \ + \ \frac{7}{150} \right)^{-\frac{2}{7}} \ \ . $$

We can construct Taylor polynomials for the powers of $ \ 3 \ $ based on the function $ \ 3^x \ $ [using $ \ \ln 3 \ \approx \ 1.1 \ $ ], and can apply the (generalized) binomial series for the other factors:

$$ 3^3 \ \cdot \ 3^{1/7} \ \cdot \left( 1 \ - \ \frac{7}{75} \right)^3 \ \left( 1 \ - \ \frac{7}{75} \right)^{\frac{1}{7}} $$ $$ \approx \ 27 \ \left( 1 \ + \ \ln 3 \cdot \frac{1}{7} \ + \ \frac{(\ln 3)^2}{2} \cdot \frac{1}{7^2} \right) \ \left( 1 \ - \ 3 \cdot \frac{7}{75}\ + \ 3 \cdot \frac{7^2}{75^2} \right) $$ $$ \cdot \left( 1 \ - \ \frac{1}{7} \cdot \frac{7}{75} \ + \ \frac{\frac{1}{7} \cdot \frac{-6}{7}}{2} \cdot \frac{7^2}{75^2} \right) \ $$

$$ \approx \ 27 \ \left( 1 \ + \ \frac{11}{10} \cdot \frac{1}{7} \ + \ \frac{6}{10} \cdot \frac{1}{7^2} \right) \ \left( 1 \ - \ \frac{7}{25}\ + \ 3 \cdot \frac{7^2}{75^2} \right) \ \left( 1 \ - \ \frac{1}{75} \ - \ \frac{3}{75^2} \right) \ \ , $$

and similarly,

$$ 3^3 \ \cdot \ 3^{-2/7} \ \cdot \left( 1 \ + \ \frac{7}{150} \right)^3 \ \left( 1 \ + \ \frac{7}{150} \right)^{-\frac{2}{7}} $$ $$ \approx \ 27 \ \left( 1 \ - \ \ln 3 \cdot \frac{2}{7} \ + \ \frac{(\ln 3)^2}{2} \cdot \frac{4}{7^2} \ - \ \frac{(\ln 3)^3}{6} \cdot \frac{8}{7^3} \right) \ \left( 1 \ + \ 3 \cdot \frac{7}{150}\ + \ 3 \cdot \frac{7^2}{150^2} \right) $$ $$ \cdot \left( 1 \ - \ \frac{2}{7} \cdot \frac{7}{150} \ + \ \frac{\frac{-2}{7} \cdot \frac{-9}{7}}{2} \cdot \frac{7^2}{150^2} \right) \ $$

$$ \approx \ 27 \ \left( 1 \ - \ \frac{11}{10} \cdot \frac{2}{7} \ + \ \frac{6}{10} \cdot \frac{4}{7^2} \ - \ \frac{4 \cdot 4}{3 \cdot 3} \cdot \ \frac{1}{7^3} \right) \ \left( 1 \ + \ \frac{7}{50}\ + \ 3 \cdot \frac{7^2}{150^2} \right) $$ $$ \cdot \ \left( 1 \ - \ \frac{1}{75} \ + \ \frac{9}{150^2} \right) \ \ . $$

In order for the series for $ \ 3^{-2/7} \ $ to remain an underestimate, it is necessary to extend the Taylor polynomial to third-degree. Simple linearizations give far too "coarse" approximations to resolve the inequality we desire clearly enough; we are also unable to obtain sufficient precision by only keeping the "linear" terms in the product. There isn't much for it other than to arrange each factor to have a single denominator, which gives us

$$ 27 \ \left[ \ \left( \frac{573}{490} \right) \ \left( \frac{4197}{5625} \right) \ \left( \frac{5547}{5625} \right) \ - \ \left( \frac{11260}{15435} \right) \ \left( \frac{25797}{22500} \right) \ \left( \frac{22209}{22500} \right) \right] \ > \ e^{\pi} \ - \ \pi^e \ \ . $$

From this point, it becomes a question of how much calculation one has the patience to do without the aid of a device; it would likely be easiest to divide out each ratio, multiply the three factors in each term, take the difference between terms, and multiply the result by $ \ 27 \ $ . In the pre-calculator era, this would not have been considered unreasonable: after all, the Ludolphine number $ ( \pi ) $ was computed to 35 decimal places over 400 years ago using polygon-approximations over the course of some years. If we avail ourselves of a calculator at this point, we see that we have

$$ 27 \ \left( \ 1.16939 \cdot 0.74613 \cdot 0.98613 \ - \ 0.72951 \cdot 1.14653 \cdot 0.98707 \right) \ \approx \ 0.9403 \ > \ e^{\pi} \ - \ \pi^e \ \ . $$

This may be compared with the first inequality we wrote, for which our result is an intended overestimate,

$$ \left( \frac{68}{25} \right)^{22/7} \ - \ \left( \frac{157}{50} \right)^{19/7} \ \approx \ 0.8901 > \ e^{\pi} \ - \ \pi^{e} \ \ . $$

$$ \ \ $$

We might well wish for something more tractable, which can be obtained by making the estimate a bit differently. If we're permitted the "number facts" $ \ e^3 \ \approx \ 20.1 \ \ , \ \ \pi^3 \ \approx \ 31 \ \ , \ \pi \ < \ \frac{22}{7} \ $ and $ \ e \ > \ \frac{19}{7} \ $ , we can proceed from

$$ \frac{201}{10} \ \cdot \ e^{1/7} \ > \ e^{3 \ + \ \frac{1}{7}} \ > \ e^{\pi} \ > \ \pi^{e} \ > \ \pi^{3 \ - \ \frac{7}{25}} \ > \ 31 \ \cdot \ \pi^{-7/25} \ > \ 31 \ \cdot \ \left( \frac{\pi}{e} \right)^{-7/25} \ \cdot \ e^{-7/25} \ \ . $$

We would like the factor $ \ \left( \frac{\pi}{e} \right)^{-7/25} \ $ to be underestimated, so we will use the overestimate

$$ \ \left( \frac{\pi}{e} \right) \ < \ \frac{22/7}{19/7} \ = \ \frac{22}{19} \ \ . $$

As above, we will use Taylor polynomials for the powers of $ \ e \ $ and the generalized binomial series to produce

$$ \frac{201}{10} \ \cdot \ \left( 1 \ + \ \frac{1}{7} \ + \ \frac{1}{2 \cdot 7^2} \right) \ > \ e^{\pi} \ > \ \pi^{e} \ > \ 31 \ \cdot \ \left( 1 \ + \ \frac{3}{19} \right)^{-7/25} \ \cdot \ \left( 1 \ - \ \frac{7}{25} \ + \ \frac{7^2}{2 \cdot 25^2} \right) $$

$$ \Rightarrow \ \ \frac{201}{10} \ \cdot \ \left( 1 \ + \ \frac{1}{7} \ + \ \frac{1}{98} \right) \ > \ e^{\pi} \ > \ \pi^{e} \ > \ 31 \ \cdot \ \left( 1 \ - \ \frac{7}{25} \cdot \frac{3}{19} \right) \ \cdot \ \left( 1 \ - \ \frac{7}{25} \ + \ \frac{49}{1250} \right) $$

$$ \Rightarrow \ \ \frac{201}{10} \ \cdot \ \frac{113}{98} \ > \ e^{\pi} \ > \ \pi^{e} \ > \ 31 \ \cdot \ \frac{454}{475} \ \cdot \ \frac{949}{1250} \ \ , $$

these ratios being obtainable by reasonably simple hand-calculation. We can make the work we face in the right-hand term of this inequality a bit less onerous by using

$$ 31 \ \cdot \ \frac{454}{475} \ \cdot \ \frac{949}{1250} \ > \ 31 \ \cdot \ \frac{450}{475} \ \cdot \ \frac{945}{1250} \ = \ 31 \ \cdot \ \frac{18}{19} \ \cdot \ \frac{189}{250} \ \ . $$

Re-arranging the inequality gives us

$$ \frac{201}{10} \ \cdot \ \frac{113}{98} \ - \ 31 \ \cdot \ \frac{18}{19} \ \cdot \ \frac{189}{250} \ = \ \frac{201 \cdot 113 \cdot 19 \cdot 25 \ - \ 31 \cdot 18 \cdot 189 \cdot 98}{980 \cdot 19 \cdot 25} \ > \ e^{\pi} \ - \ \pi^{e} \ \ . $$

While this looks a bit daunting, it is considerably easier to compute by hand than the earlier estimate: a little work with binomials simplifies some of the multiplicaton and the rest is not too bad to work out "long-hand". Hence,

$$ \frac{10,788,675 \ - \ 10,335,276}{465,500} \ = \ \frac{453,399}{465,500} \ > \ e^{\pi} \ - \ \pi^{e} \ \ . $$

So we again establish the desired inequality, this time with more handiwork. We can make use of the calculator (or divide long-hand if we *really* want to remain "pure") to find that this ratio is about 0.9740 .

I will admit to a bit of guidance from a calculator to make adequate selections for the rational approximations in the interests of saving some time. It would take *rather* longer than to refine the numbers used in this investigation if one had to work *entirely* without "machine assistance".