Quick induction proof I am trying to prove $n^3<n!$ for all integers $n\geq 6.$ It would be trivial to do this by induction if $(n+1)^3<(n+1)n^3$ holds. I looked this up, and I found this is true for integers $n\geq 3.$ However, does anybody know of any more insightful method of proof for either statement, ie. not induction? Thank you. 

You can avoid induction if you allow division in the following way: 
Let $n\geq 6$. Notice that
\begin{align*}
n!&=n(n-1)(n-2)(n-3)(n-4)(n-5)!\\
&\geq n(n-1)(n-2)3\cdot 2\cdot 1\tag{1}\\&=6n(n-1)(n-2).
\end{align*}
Now, we can make the following evaluation:
\begin{align*}
\dfrac{n^3}{n(n-1)(n-2)}=\dfrac{n}{n-1}\dfrac{n}{n-2}.\tag{2}
\end{align*}
It's easily proven that, for a given $k$, the function $f_k:n\mapsto\dfrac{n}{n-k}$ (defined for $n>k$) is decreasing. In particular, since $n\geq 6$, then
\begin{align*}
\dfrac{n}{n-1}=f_1(n)\leq f_1(6)=\dfrac{6}{5}\quad\text{and}\quad\dfrac{n}{n-2}=f_2(n)\leq f_2(6)=\dfrac{6}{4},\tag{3}
\end{align*}
so, using (2) and (3), we obtain
$$\dfrac{n^3}{n(n-1)(n-2)}\leq\dfrac{6}{5}\dfrac{6}{4}=\dfrac{36}{20}<6,\tag{4}$$
hence, by (4) and (1),
$$n^3\leq6n(n-1)(n-2)\leq n!$$
