How to prove that for all natural numbers, $4^n > n^3$? This is a problem set I have, it's not a homework but it's very important practice...
Send me some hints please, I don't want an answer I need to get it by myself but I'm failing miserably... 
The problem is:

Prove that for all natural numbers $n$,  $4^n > n^3$

 A: Let's see: We have that $2^n>n$ for all $n$, and $2^n>n^2$ for $n\ge 5$. So $4^n=2^n2^n>n^3$ for $n\ge 5$, and we just need to prove the cases $n\le 4$ by hand, and this is straightforward. 
A: Your qn in the title and the body are not the same. Taking the version in the title as your question here is a suggestion to attack:
Take the ratio of consecutive terms. For the LHS  the ratio 
is $4^{n+1}/4^n = 4$. That is every time the LHS gets magnified by a factor of 4.  For the RHS the ratio is $\frac{ (n+1)^3}{n^3}$ which keeps getting smaller;  $(3/2)^3=27/8$ is already less than 4. So you can see that LHS will be bigger.
A: I made it for every natural $n\geq 3$, $4^n>n^3$.
For induction in $n$. First, for $n=3$ is ok : $4^3>3^3$. Therefore, it is the basis of induction.
Now suppose that the inequality is true to a certain $n=k$, i.e., $4^k>k^3$. So as $k\geq 3$, it follows that $k^2\geq 9>3$ and $k^3\geq 9>1$, then
$$4^{k+1}=4^k\cdot 4>k^3\cdot 4=k^3+k^3+k^3+k^3>k^3+1 +k^2\cdot k+k\cdot k^2>$$
$$>k^3+1+3k^2+3k=(k+1)^3. $$
Therefore, the result follows by induction.
A: Induction is clearly enough, as pointed by @user131755. 
If you like another approach, consider the function $f:\mathbb{R}^+\to\mathbb{R}^+$ given by $f(x)=x^3 e^{-\log(4) x}$. Its derivative is:
$$f'(x) = \left(3x^2-\log(4)x^3\right)e^{-\log(4) x},$$
so $f(x)$ has a maximum in $x=\frac{3}{\log 4}$ and 
$$ f(x) \leq f\left(\frac{3}{\log 4}\right)=\left(\frac{3}{e\log 4}\right)^3\leq\left(\frac{9}{10}\right)^3<1 $$
since $e>\frac{5}{2}$ and $\log(2)>\frac{2}{3}$. $f(x)<1$ just gives $4^x>x^3$ for any $x>0$.

In alternative, for any $n\geq 3$ by the binomial theorem we have:
$$ 4^{n} = \frac{1}{4}(3+1)^{n+1} > \frac{27}{4}\binom{n+1}{3} = \frac{9}{8}(n^3-n)=n^3+\frac{n(n+3)(n-3)}{8}\geq n^3.$$
A: Try induction. $4^n=4 \cdot 4^{n-1}$ and the binomial theorem might come in handy.
