Factorial Greater than Cube for n Greater than 5/Proof 2

Theorem

Let $n \in \Z$ be an integer such that $n > 5$.

Then $n! > n^3$.

For $n > 5$, notice that the following inequalities hold:

$2 \paren {n - 1} = 2 n - 2 > n + 5 - 2 > n$

$3 \paren {n - 2} = 3 n - 6 > n + 10 - 6 > n$

And thus:

$\ds n!$

$\ge$

$\ds n \paren {n - 1} \paren {n - 2} \paren {3!}$

$\ds $

$=$

$\ds n \paren {2 \paren {n - 1} } \paren {3 \paren {n - 2} }$

$\ds $

$>$

$\ds n \paren n \paren n$

$\ds $

$=$

$\ds n^3$

$\blacksquare$