Proving by induction: $2^n > n^3 $ for any natural number $n > 9$ 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???

If $P(n): 2^n>n^3$
$n=10, P(10): 2^{10}=1024$ and $10^3=1000$
Let $P(n)$ is true for $n=m\implies 2^m>m^3$
Now, $P(m+1): 2^{m+1}=2\cdot 2^m>2m^3$ which we need to be $>(m+1)^3$
$\implies 2>\left(1+\frac1m\right)^3 $
If $m=2, \left(1+\frac1m\right)^3=\frac{27}8>2$
If $m=3, \left(1+\frac1m\right)^3=\frac{64}{27}>2$
If $m=4, \left(1+\frac1m\right)^3=\frac{125}{64}<2$
$\implies  2>\left(1+\frac1m\right)^3$ for $m=4$
and $\left(1+\frac1m\right)^3>\left(1+\frac1{m+1}\right)^3\implies 2>\left(1+\frac1m\right)^3$ for $m\ge 4$
