> 
> 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](https://en.wikipedia.org/wiki/Taylor%27s_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](https://en.wikipedia.org/wiki/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 $(1)(3)(4)$, 
$$e^{\pi}-{\pi}^e\lt \frac{e\pi(\pi-e)^2}{2}\cdot\frac{1}{4-\pi}\lt\frac{3\times\frac{22}{7}\left(\frac{22}{7}-2.718\right)^2}{2}\cdot\frac{1}{4-\frac{22}{7}}\lt 0.993\lt 1$$