Prove for any $\ n \ge 1 $
$$\ e^{n^2} \ge n^n $$
my attempt:
true for $\ n = 1 $ then for $\ n+ 1 $ :
$\ e^{(n+1)^2} = e^{n^2 + 2n + 1} = e^{n^2} \cdot e^{2n} \cdot e $
$\ ( n+1)^{n+1} = (n+1)^n\cdot (n+1) \le (n+n)^n\cdot(n+1) = (2n)^n \cdot(n+1) = 2^n \cdot n^n \cdot(n+1) \le 2^n \cdot e^{n^2} \cdot(n+1) $
from here it's obvious $\ e^{2n} \ge 2^n $ but I can't find a way to get rid of $\ n+1 $ and $\ e $?
hint
Taking the $n^{\text{th}} $ root, it is much easier to prove that $$(\forall n\ge 1)\; e^n\ge n$$
Let $ n\ge 1$ such that $ e^n\ge n.$
$$e^n\ge n\implies e^{n+1}\ge en$$ $$\implies e^{n+1}\ge 2n$$
$$\implies e^{n+1}\ge n+n$$ $$\implies e^{n+1}\ge n+1$$
Cantor's inequality $2^n > n$ implies $e^n > 2^n > n$ for any $n$.
Then just raise it to $n$-th power.
$e^{n^{2}} \geq n^n$ $\Leftrightarrow $$e^{n^{2}} \geq e^{n\ln(n)} $ $\Leftrightarrow $ $ n^2 \geq n\ln(n) $ $\Leftrightarrow $ $n\geq \ln(n) $ that is correct for all $n>1$
You can continue your work by taking $\ln$ of both sides and we have $2n+1 \geq n\ln2 + \ln(n+1) \geq n/2 + n+1=3n/2 + 1$. Then we get that $n/2 \geq 0$ with reversible steps.
