Proving that $e^{\pi}-{\pi}^e\lt 1$ without using a calculator 
Prove that $e^{\pi}-{\pi}^e\lt 1$ without using a calculator. 

I did in the following way. Are there other ways?
Proof : Let $f(x)=e\pi\frac{\ln x}{x}$. Then, 
$$e^{\pi}-{\pi}^e=e^{f(e)}-{e}^{f(\pi)}\tag1$$
Now,
$$f'(x)=\frac{e\pi(1-\ln x)}{x^2},\quad f''(x)=\frac{e\pi (2\ln x-3)}{x^3},\quad f'''(x)=\frac{e\pi (11-6\ln x)}{x^4}.$$
Since $f'(x)\lt 0$ for $e\lt x\lt\pi$, one has $f(e)\gt f(\pi)$. By Taylor's theorem, there exists a point $c$ in $(e,\pi)$ such that 
$$\begin{align}f(\pi)&=f(e)+(\pi-e)f'(e)+\frac{(\pi-e)^2}{2}f''(c)\\&\gt f(e)+(\pi-e)\cdot 0+\frac{(\pi-e)^2}{2}\cdot \frac{e\pi(2\ln e-3)}{e^3}\\&=f(e)-\frac{\pi(\pi-e)^2}{2e^2}\tag2\end{align}$$
because $f'''(x)\gt 0\ (e\lt x\lt \pi)$ implies $f''(c)\gt f''(e)$.
By the mean value theorem and $(2)$, 
$$e^{f(e)}-e^{f(\pi)}\lt (f(e)-f(\pi))e^{f(e)}\lt \frac{\pi (\pi-e)^2}{2e^2}\cdot e^{\pi}=\frac{e\pi(\pi-e)^2}{2}\cdot e^{\pi-3}\tag3$$
Since $e^x\lt \frac{1}{1-x}\ (0\lt x\lt 1)$ and $0\lt \pi-3\lt 1$, 
$$e^{\pi-3}\lt\frac{1}{4-\pi}\tag4$$
From $(1)(3)(4)$, 
$$e^{\pi}-{\pi}^e\lt \frac{e\pi(\pi-e)^2}{2}\cdot\frac{1}{4-\pi}\lt\frac{3\times\frac{22}{7}\left(\frac{22}{7}-2.718\right)^2}{2}\cdot\frac{1}{4-\frac{22}{7}}\lt 0.993\lt 1$$
 A: Consider $f(x,y) = x^y - y^x$ where $x, y \approx 3$.  Increasing $x$ should bring $f$ down and increasing $y$ should bring $f$ up.  In general, without any real knowledge of $f$:
$$ 1 > f(x+ \epsilon_1, y + \epsilon_2) \approx 
 f(x, y ) + \epsilon_1 \frac{\partial f}{\partial x}  + \epsilon_2 \frac{\partial f}{\partial y}
 $$
where the first inequality is something we are trying to prove for $x = e \approx 2.718$ and $y = \pi \approx 3.141$.

Let's try $x = \frac{11}{4}$ and $y = \frac{13}{4}$.   Then we have taken off too much slack:
$$ \left( \frac{11}{4} \right)^{\frac{13}{4}} - \left( \frac{13}{4} \right)^{\frac{11}{4}} \approx 1.214$$
Using the continued fractions of $e$ and $\pi  $ does give less than one and check with a calculator:
$$ \left( \frac{8}{3} \right)^{\frac{22}{7}} - \left( \frac{22}{7} \right)^{\frac{8}{3}} \approx 0.621$$
Even your proof requires a calculator in the last step... but we did not use the exact values of $\pi, e$.  

How good is this approximation?  Somehow we should compute how quickly $f$ changes with $x$ and $y$:
$$ \frac{\partial f}{\partial x} = y \, x^{y-1} - (\ln y ) y^x \approx 3 \times 3^{3-1} - (\ln 3) 3^3 \approx -27(\ln \frac{e}{3}) \approx -2.7 $$
The $y$ partial derivative is similar, except it is positive instead of negative.  

It can be proven, the continued fraction error of $e$ and $\pi$ are inversely proportional to the denominators:
$$ \left| e - \frac{8}{3} \right| < \frac{1}{3^2} \text{ and }  \left| \pi - \frac{22}{7} \right| < \frac{1}{7^2}$$
These would be our estimates of $\epsilon_1$ and $\epsilon_2$.  Then using  a tiny bit of multivariable calculus:
$$ \epsilon_1 \frac{\partial f}{\partial x}  + \epsilon_2 \frac{\partial f}{\partial y}
\approx \frac{1}{3^2} \times 2.7 + \frac{1}{7^2} \times (-2.7) < 0.355 $$
This should be enough to establish $|f(\pi,e)| < 1$.

This used a calculator in many places but not the exact value of the constants $\pi$ and $e$.
A: Let $x>0$. Consider $e^{e+x}-(e+x)^e=e^e(e^x-(1+\frac xe)^e)=e^e(A-B)$.
Now $z\ge \log(1+z)\ge z-\frac{z^2}2$, so $0\le A-B=A(1-\frac BA)\le A\log(A/B)\le A\frac{x^2}{2e}$. Hence, for $x=\pi-e$, we get
$$
0\le e^{\pi}-\pi^e\le e^{\pi-1}(\pi-e)^2/2\,.
$$
Now we use the trivial inequality $e\ge 2+\frac 12+\frac 16+\frac 1{24}=2+\frac{17}{24}>2.7$ ($170>168$ if you want to check it mentally) and the much less trivial inequality $\pi<3.15$, for which I have no paper-free proof, to get $(\pi-e)^2\le 0.45^2=0.2025<\frac 29$ (the two-digit numbers ending in $5$ can be squared without paper). So, all we need to show now is that $\log 9=2\log 3\ge 2.15$ or $-\log \frac 13>1.075$. However, writing the first 6 terms of the Taylor series, we see that this logarithm is at least 
$$
\frac 23+\frac 29+\frac 8{81}+\frac 4{81}+\frac{32}{5\cdot 3\cdot 81}
+\frac{64}{6\cdot 9\cdot 81}
\ge 1+\frac 3{81}+\frac 2{81}+\frac{2}{15\cdot 81}+\frac 1{81}
\\
>1+\frac 6{81}+\frac 1{8\cdot 81}>1+\frac 6{81}+\frac 6{80\cdot 81}=1+\frac 6{80}=1.075
$$
So, if we define $\pi$ to be a number between $e$ and $3.15$, the whole computation (assuming decent memory) requires no pen and paper, not to mention calculators. However, $\pi$ is not defined that way, so the inequality $\pi<3.15$ is a gaping flaw in this argument :-(.
A: Take the roots against $\pi e$, to get $e^{1/e}$ v $\pi^{1/\pi}$.  These are instances if $x^{1/x}$ for $x=e$ and $x=\pi$.
For integers, we see that 9>8 and 2^4=4^2=16 imply that $\sqrt[3]{3} > \sqrt[4]{4} = \sqrt{2}$, from the sixth and eighth roots respectively.
Put $y=\sqrt[x]{x}$, whence $x = y^x$. We note that for values greater than 1, there are two solutions, the second is $z=y^z$.  We now suppose z>x, and consider the case w>z.  Where w>z, then $w<y^w$, which means that the value of y for w is less than that for z.  
In the region where 1 < w < x  we note that $y^x > y^w$, still, and that $\sqrt[w]{w} < \sqrt[x]{x} < L$
If now we note that $\pi > 3 > e$, and if we find a corresponding value for 3, then the range from 3 to 3' will contain either pi or e.
We now seek a value r for which $r^3 = 3^r$.  Put r=2.5, gives r^3 = 125/8 = 15.625, and $9\sqrt{3}=1.732*9$, at 15.588.  This tells us that r<2.5, and we get $r=3'< 2.5 < e < 3 < \pi$. 
Since this is a convergence of $x$ and $z$ converge on some limit $L$, we can use brackets of the form of ( and ) to show this convergance, viz
$ 3' (2.5  ( e )) 3 ) \pi$, and thus from $3' ( e ) 3 ) \pi$, $e^\pi > \pi ^e$.
