Proof by Induction $3^n > n^3$ I am trying to prove the following, however I'm stuck at the Induction hypothesis

Prove by induction that, for all integers $n$, if $n\geq 5$, then $3^n>n^3$

What I have Done:
Base Case:
$n = 5$
$3^5 > 5^3$
$243 > 125$ so TRUE
Assume True for $n = k$
$3^k > k^3$ must be TRUE
Should be TRUE for $n = k+1$
$3^k + 3^{k+1} > k^3 + 3^{k+1}$
Im a little stuck here. Any help would be appreciated!

By induction, you assume that for $ n = k,
3^k > k^3$ must be true. We want to show $ 3^{k+1} > (k+1)^3$.
Thus,
$3^{k+1} = 3^k\cdot 3 >3\cdot k^3$, by your induction hypothesis.
So we want to show $ 3^{k+1} > (k+1)^3$, but have $3^{k+1} > 3k^3$.
So just show $3k^3 > (k+1)^3$ for $k>5$. We have
$(k+1)^3 = k^3+3k^2+3k +1 < k^3+k^3+k^3 = 3k^3$.
Therefore,$3^{k+1} > 3k^3 > (k+1)^3 \Longrightarrow 3^{k+1} > (k+1)^3,$ 
and we are done.
