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???

For another way just using $n>9$, 
note that when $n=10$, $2^n = 1024 > 1000 = n^3$. Now suppose that $2^n>n^3$ for $n>9$. Then, 
$\begin{align*}
2^{n+1} &= 2\cdot2^n \\
&>2n^3 \\
&= n^3 +n^3 \\
&> n^3 + 9n^2 \\
&= n^3 + 3n^2 + 6n^2 \\
&>n^3 + 3n^2 +54n \\ 
&=n^3+3n^2+3n +51n\\
&>n^3+3n^2+3n+1 \\
&= (n+1)^3.
\end{align*}$
