Prove $3^n \ge n^3$ by induction Yep, prove $3^n \ge n^3$, $n \in \mathbb{N}$.
I can do this myself, but can't figure out any kind of "beautiful" way to do it. 
The way I do it is:
Assume $3^n \ge n^3$
Now, 
$(n+1)^3 = n^3 + 3n^2 + 3n + 1$,
and $\forall{} n \ge 3$,
$3n^2 \le n^3, \,\, 3n + 1 \le n^3$ 
Which finally gives $(n+1)^3 \le 3n^3 \le 3^{n+1}$ by our assumption.
Now just test by hand for n=1,2,3 and the rest follows by induction.
Anyone got anything simpler?

Here is another argument, but it's not necessarily simpler than yours:
1) If $n=1$, $3^1=3\ge1=1^3$;  if $n=2$, $3^2=9\ge8=2^3$; and if $n=3$, $3^3=3^3$.
2) Now assume that $3^n\ge n^3$ for some integer $n\ge3$.
Then $\displaystyle\frac{1}{n}\le\frac{1}{3}$, so $\displaystyle\frac{(n+1)^3}{n^3}=\big(1+\frac{1}{n}\big)^3\le\big(\frac{4}{3}\big)^3=\frac{64}{27}\le3$, so 
$\;\;\;\;\;\;\;\;3^{n+1}=3(3^n)\ge3n^3\ge(n+1)^3$.
