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???
 A: For your "subproof": 
Try proof by induction (another induction!) for $k \geq 7$
$$k^3 > 3k^2 + 3k + 1$$
And you may find it useful to note that $k\leq k^2, 1\leq k^2$
$$3k^2 + 3k + 1 \leq 3(k^2) + 3(k^2) + 1(k^2) = 7k^2 \leq k^3 \quad\text{when}??$$
A: Here's another way. Suppose $k>9$. Then:
$$ \begin{align*}
k^3&=(k)k^2\\
&>9k^2\\
&=3k^2+3k^2+3k^2\\
&=3k^2+3(k)k+3(k)^2\\
&>3k^2+3(9)k+3(9)^2\\
&=3k^2+27k+243\\
&>3k^2+3k+1\\
\end{align*}$$
A: 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 $m\ge 4$
A: Hint:  You can use that if $k\ge10$, then $k^3\ge 10k^2=3k^2+7k^2$.
(Another approach would be to use that $\frac{(k+1)^3}{k^3}=(1+\frac{1}{k})^3\le
(11/10)^3<2$ for $k\ge10$.
A: 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 $n \ge 10$.
A: 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, 
$\begin{align*}
2^{n+1} &= 2\cdot2^n \\
&>2n^3 \\
&= n^3 +n^3 \\
&> n^3 + 9n^2 \\
&= n^3 + 3n^2 + 6n^2 \\
&>n^3 + 3n^2 +54n \\ 
&=n^3+3n^2+3n +51n\\
&>n^3+3n^2+3n+1 \\
&= (n+1)^3.
\end{align*}$
