TITLE: How prove $\pi^2>2^\pi$
QUESTION [5 upvotes]: 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

REPLY [3 votes]: This is similar to this problem, asking which of $\pi^3$ and $3^\pi$ is bigger.  It can be deduced from the inequalities
$$3\lt\pi\lt{22\over7}$$
Specifically,
$$2^{11}=2048\lt2187=3^7\implies2^{22/7}\lt3^2\implies 2^\pi\lt\pi^2$$