How prove $\pi^2>2^\pi$ show that
$$\pi^2>2^\pi$$
I use computer found 
$$\pi^2-2^\pi\approx 1.044\cdots,$$
can  see this
I know 
$$\Longleftrightarrow \dfrac{\ln{\pi}}{\pi}>\dfrac{\ln{2}}{2}$$
so let
$$f(x)=\dfrac{\ln{x}}{x}$$
so
$$f'(x)=\dfrac{1-\ln{x}}{x^2}=0,x=e$$
so $f(x)$ is Strictly increasing
 on $(2,e)$, and is Strictly decreasing
 on $(e,3) $
so I can't know $f(2)$ and $f(\pi)$ which is bigger?
maybe this problem exsit have easy methods by hand

For $n>e$ $f(x)$-decreasing so $ln(\pi)/\pi>ln(4)/4$ but $ln(4)/4=ln(2)/2$
