# Factorial Greater than Cube for n Greater than 5

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## Contents

## 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:

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

$\blacksquare$

## Sources

- 1980: David M. Burton:
*Elementary Number Theory*(revised ed.) ... (previous) ... (next): Chapter $1$: Some Preliminary Considerations: $1.1$ Mathematical Induction: Problems $1.1$: $6$