Q:

Prove using induction that for $n \ge 4$ the inequality $n^3 < 3^n$ holds.

I need to prove using induction that for $n \ge 4$ the inequality $n^3 < 3^n$ holds. I know that this problem has already been posted here but I managed to solve this in a simpler way than those offered by other proposers and so I decided to post it again to ask if it's actually correct.
(1) I check the basic case: $64 < 81 $ and so it works 
(2) I assume that the statement is is true for all numbers greater than or equal to $k$ 
(3) for k+1:
$$(k+1)^3=k^3+3k^2+3k+1 < k^3 + 3k^2 + 3k + k^3 < (*) $$
Now, I'd like to show that $3k^2 +3k < k^3 \Rightarrow k^2 - 3k  -3 > 0 $Since this is an increasing parabola, this inequality will be satisfied for every $k$ greater than $n$, and $3 < n < 4 $ therefore, for $n = 4 n^3 > 3n^2 + 3n$

Therefore
$$(*) < k^3 + k^3+k^3 = 3k^3 = 3(3^k) = 3^{k+1}$$
QED 
What do you think of this solution?

A:

The middle part is a little weird.  What you need to do is to show that if $k>4$ then $3k^2+3k < k^3$.  From what you have written it looks as if you're going the other way ... and end up with $k < 4$ rather than starting with $k>4$ ... or was this supposed to be some kind of contraposition?

