TITLE: Prove that $n! > n^{3}$ for every integer $n \ge 6$ using induction
QUESTION [5 upvotes]: Proof.
Base case. When $n = 6$, LHS = $6! = 720$ and RHS = $6^{3} =  216$. Thus LHS > RHS.
Inductive step. Assume that $m! > m^{3}$ for some fixed integer $m$ such that $m \ge 6$. We must show that it also holds true for $(m+1)$, i.e, $(m+1)! > (m+1)^{3}$. Observe that 
\begin{align*}
(m+1)! &= (m+1) \cdot m! \\
&> (m+1) \cdot m^{3} \text{$\quad$(by the inductive hypothesis)}\\
\end{align*}
(still unfinished)

I know it boils down to showing that $(m+1) \cdot m^{3} > (m+1)^{3}$ but I'm unsure on how to proceed.

REPLY [3 votes]: If you want to show $(m+1)\cdot m^3>(m+1)^3$, this is equivalent to
$$m^3>(m+1)^2.$$
As $m>2$, this is implied by
$$2m^2>(m+1)^2,$$
which is equivalent to
$$m^2>2m+1$$
or
$$(m-1)^2>2,$$
which is true if $m\geq 3$.
(Note that if you were to write this up it would be clearer if you started from $(m-1)^2>2$, which you know is true, and worked back to $m^3>(m+1)^2$ -- this writeup serves to help make clear how one would come up with this.)