Prove through induction that $3^n > n^3$ for $n \geq 4$ I'm new to induction and have not done induction with inequalities before, so I get stuck at proving after the 3rd step.
The question is:  

Use induction to show that $3^n > n^3$ for $n \geq 4$.

I have so far:
Step 1:
Prove for $n=4$ (since question states this)
$3^4 > 4^3$
$81 > 64  $
which is true 
Step 2:
Assume true for $n=k$

$3^k > k^3$

Step 3:
Prove for $n = k+1$  

$3^{k+1} > (k+1)^3$

Here I expand to:  

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

However I have no idea how to prove this.
Thanks for any help given

$3^n > n^3$ for $n\geq 4$
Base case:
If $n=4$, we get 
$\text{LHS}=3^4=81$
$\text{RHS}=4^3=64$.
We can see that $\text{LHS} > \text{RHS}$.
Let us assume the above hypothesis holds true for some $n=k$ such that $k\geq 4$.
Therefore,
$$3^k > k^3$$
$$3(3^k) > 3k^3$$
$$3^{k+1} > k^3 + k^3 + k^3 > k^3 + 3k^2 + 3k^2 > k^3 + 3k^2 + k^2 + k^2 + k^2  > k^3 + 3k^2 + 3k + 3k + 3k > k^3 + 3k^2 + 3k + 1 = (k+1)^3.$$
