# 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:

 $\displaystyle 1!$ $=$ $\displaystyle 1$ $\displaystyle$ $=$ $\displaystyle 1^3$ $\displaystyle 2!$ $=$ $\displaystyle 2$ $\displaystyle$ $<$ $\displaystyle 8$ $\displaystyle$ $=$ $\displaystyle 2^3$ $\displaystyle 3!$ $=$ $\displaystyle 6$ $\displaystyle$ $<$ $\displaystyle 27$ $\displaystyle$ $=$ $\displaystyle 3^3$ $\displaystyle 4!$ $=$ $\displaystyle 24$ $\displaystyle$ $<$ $\displaystyle 64$ $\displaystyle$ $=$ $\displaystyle 4^3$ $\displaystyle 5!$ $=$ $\displaystyle 120$ $\displaystyle$ $<$ $\displaystyle 125$ $\displaystyle$ $=$ $\displaystyle 5^3$

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:

 $\displaystyle 6!$ $=$ $\displaystyle 720$ $\displaystyle$ $>$ $\displaystyle 216$ $\displaystyle$ $=$ $\displaystyle 6^3$

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:

 $\displaystyle \paren {k + 1}!$ $=$ $\displaystyle \paren {k + 1} \times k!$ $\displaystyle$ $>$ $\displaystyle \paren {k + 1} \times k^3$ Induction Hypothesis $\displaystyle$ $=$ $\displaystyle \paren {k - 3} \times k^3 + 3 \times k^3 + k^3$ $\displaystyle$ $>$ $\displaystyle 1 + 3 \times k + 3 \times k^2 + k^3$ $\displaystyle$ $=$ $\displaystyle \paren {k + 1}^3$

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

Therefore:

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

$\blacksquare$