Factorial Greater than Cube for n Greater than 5

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

Then $n! > n^3$.

Proof
We note that:

The proof then proceeds by induction.

For all $n \in \Z_{\ge 6}$, let $\map P n$ be the proposition:
 * $n! > n^3$

Basis for the Induction
$\map P 6$ is the case:

Thus $\map P 6$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown that if $\map P k$ is true, where $k \ge 6$, then it logically follows that $\map P {k + 1}$ is true.

So this is the induction hypothesis:
 * $k! > k^3$

from which it is to be shown that:
 * $\paren {k + 1}! > \paren {k + 1}^3$

Induction Step
This is the induction step:

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\forall n \in \Z_{>5}: n! > n^3$