For what natural $n$ does $3^n > n^3$ hold true? Prove by induction 
For what natural $n$ does $3^n > n^3$ hold true? 

I figured that it holds true for all $n$ except $n = 3$. I am not sure how to prove it by induction. I proved it by $p(k) \implies p(k+1)$ but that doesn't show that $n \neq 3$ 
 A: You can prove it for $n \ge 4$ by inductive hypothesis. Set the base case to be $4$ and then prove that $p(k) \implies p(k+1)$. For the other three cases you can give a case-by-case proof, checking each case seperately and hence deduce a conclusion.
A: For $n=4$ we have
$$3^n> n^3$$ is true.
Let $$3^n> n^3$$ for all $n>3$.
Thus, $$3^{n+1}=3\cdot3^n>3n^3$$ and it's enough to prove that
$$3n^3>(n+1)^3$$ or
$$\sqrt[3]3n> n+1,$$ which is true for $n>3.$
A: The very final step would be:
\begin{align*}
3^{n+1}&=3\cdot 3^{n}\\
&> 3n^{3}
\end{align*} and we have 
\begin{align*}
3n^{3}-(n+1)^{3}&=2n^{3}-3n^{2}-3n-1\\
&=n^{3}-3n^{2}+n^{3}-3n-1\\
&\geq n^{2}(n-3)+n^{3}-3n-n\\
&=n^{2}(n-3)+n(n+2)(n-2)\\
&>3\cdot5\cdot 1\\
&>0
\end{align*}
since $n> 3$. As a matter of simple checking, one should starts from $4$.
A: We have $\;3^n>n^3\;$ for $n=4$.
So suppose we have $\;3^n>n^3\iff u_n=\dfrac{3^n}{n^3}>1\;$ for some $n\ge 4$. To show $u_n>1$ for all $n\ge 4$, it suffices to show $\;\dfrac{u_{n+1}}{u_n}\ge 1$. 
Note that, by Bernoulli's inequality,
$$\frac{u_{n+1}}{u_n}=\frac{3^{n+1}}{3^n}\frac{n^3}{(n+1)^3}=3\Bigl(1-\frac1{n+1}\Bigr)^3\ge 3\Bigl(1-\frac3{n+1}\Bigr)=\frac{3(n-2)}{n+1}$$
and $\;\dfrac{3(n-2)}{n+1}\ge 1\iff 2n\ge 7$, which is true if $n\ge 4$.
