Proof by induction: inequality $n! > n^3$ for $n > 5$ I'm given a inequality as such: $n! > n^3$
Where n > 5, 
I've done this so far:
BC: n = 6, 6! > 720 (Works)
IH: let n = k, we have that: $k! > k^3$
IS: try n = k+1, (I'm told to only work from one side)
So I have (k+1)!, but I'm not sure where to go from here.
I've been told that writing out: $(k+1)! > (k+1)^3$ is a fallacy, because I can't sub in k+1 into both sides, but rather prove from one side only.  
Any Help on how to continue from here, would be much appreciated. 
 A: Note that it is equivalent to prove that $n!/n^{3} > 1$ for all $n \geq 6$. You have proved for $n = 6$; suppose $n \geq 6$ is such that $n! > n^{3}$; then 
$$
\frac{(n+1)!}{(n+1)^{3}} > \frac{n^{3}(n+1)}{(n+1)^{3}} = \frac{n^{3}}{(n+1)^{2}} = \frac{n^{3}}{n^{2}+2n+1} > \frac{n^{3}}{4n^{2}} = \frac{n}{4} > 1.
$$
You know how to conclude.
A: You have $(k+1)!=(k+1)k!>(k+1)k^3$ by the induction hypothesis. Now it suffices to show that if $k>5$ then $k^3 > (k+1)^2$. To do that, we write $(k+1)^2 = k^2 +2k+1 < k^2 + kk + k^2 = 3k^2 < kk^2 = k^3$, since we're assuming $k>5$.
A: Suppose
$n \ge 6$
and
$n! > n^3$.
You want to show that
this implies that
$(n+1)!
> (n+1)^3
$.
Since
$n! > n^3$
and
$(n+1)! 
=(n+1)n!
$,
we have
$(n+1)!
> (n+1)n^3
$.
So,
we are done if
we can show that
$(n+1)n^3 \ge (n+1)^3
$.
But,
dividing by $n+1$,
 this is equivalent to
$n^3
\ge (n+1)^2
$.
There are a variety of ways 
to prove this -
you can even use induction.
Here is a direct way:
Since $n \ge 6$,
$(n+1)^2
=n^2(1+\frac1{n})^2
\le n^2(1+\frac1{6})^2
\le n^2\frac{49}{36}
< 2n^2
$.
But
$n^3 \ge 6n^2
> 2n^2
> (n+1)^2
$
and we are done.
