TITLE: Proving by induction: $2^n > n^3 $ for any natural number $n > 9$
QUESTION [6 upvotes]: I need to prove that $$ 2^n > n^3\quad \forall n\in \mathbb N, \;n>9.$$
Now that is actually very easy if we prove it for real numbers using calculus. But I need a proof that uses mathematical induction. 
I tried the problem for a long time, but got stuck at one step - I have to prove that:

$$ k^3 > 3k^2 + 3k + 1 $$

Hints???

REPLY [2 votes]: Your problem,
$2^n > n^3$,
 is equivalent to
$n < 2^{n/3}$.
Suppose
$n < 2^{n/3}$.
Then
$2^{(n+1)/3}
=2^{1/3}2^{n/3}
> n 2^{1/3}
$
and
$n 2^{1/3} > n+1
\iff n (2^{1/3}-1) > 1
\iff n > \frac1{2^{1/3}-1}
\iff n > 3.847...
$.
So, if
$n \ge 4$
and
$n^3 < 2^n$,
then
$(n+1)^3 < 2^{n+1}$.
Since
$1000 = 10^3 < 2^{10} = 1024$,
$n^2 < 2^n$
for $n \ge 10$.