prove $\ e^{n^2} \ge n^n $ for every n by induction 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 $?

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.
