Prove by induction on $n\geq k$ that $n^3 \lt 3^n$. What is the value $k$? Prove by induction on $n\geq k$ that $n^3 \lt 3^n$. What is the value $k$?
It looks like $k$ should be $0$ after trying random values but obviously that is a terrible way of doing things.
Proof:
P(n): n^3<3^n

Assume k holds
Prove k+1
P(k+1): (k+1)^3<3^(k+1)

At this point, I am not entirely sure where to go. Any advice would be great.
Thanks,

First we compute a little, systematically. We have
$0^3\lt 3^0$;
$1^3 \lt 3^1$;
$2^3\lt 3^2$;
$3^3=3^3$, so there inequality $n^3\lt 3^n$ is not true at $n=3$;
$4^3=64$, $3^4=81$, so $4^3\lt 3^4$;
$5^3=125$, and $3^5=243$;
$6^3=216$, and $3^6=729$;
$7^3-343$, and $3^7=2187$.
If this were a horse race, it looks as if the horse $3^n$ is now well ahead of the horse $n^3$, and is getting further and further ahead as time goes on. 
We will prove that $n^3\lt 3^n$ if $n\ge 4$.
The result is certainly true at $n=4$. We show that for any $k\ge 4$, if $k^3\lt 3^k$, then $(k+1)^3\lt 3^{k+1}$. There are several ways to do this. We mention a couple.
(i) We have $3^{k+1}=3\cdot 3^k$. By the induction assumption, $3^k\gt k^3$. So it is enough to show that $3k^3\ge (k+1)^3$, or equivalently that $3\ge \left(1+\frac{1}{k}\right)^3$. Note that $\left(1+\frac{1}{4}\right)^3\lt 3$, and if $k\gt 4$, then $\left(1+\frac{1}{k}\right)^3\lt  \left(1+\frac{1}{4}\right)^3$. This completes the proof.  The idea behind this proof is that if we increment $n$ by $1$, then $3^n$ jumps by a factor of $3$, but $n^3$ only jumps by a factor of $\left(\frac{n+1}{n}\right)^3$, which very soon is $\lt 3$.
(ii) We want to show that if $k\ge 4$, then $3k^3\ge (k+1)^3$, or, expanding, that $2k^3\gt 3k^2+3k+1$. Note that if $k\ge 4$, then $2k^3\ge 8k^2$. But $3k^2+3k+1\lt 3k^2+3k^2+k^2=7k^2$. 
Alternately, we could look at the function $2x^3-3x^2-3x-1$, and use techniques from the calculus or elsewhere to find out when it is positive. 
