Factorial Greater than Cube for n Greater than 5/Proof 2
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Theorem
Let $n \in \Z$ be an integer such that $n > 5$.
Then $n! > n^3$.
Proof
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:
\(\ds n!\) | \(\ge\) | \(\ds n \paren {n - 1} \paren {n - 2} \paren {3!}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds n \paren {2 \paren {n - 1} } \paren {3 \paren {n - 2} }\) | ||||||||||||
\(\ds \) | \(>\) | \(\ds n \paren n \paren n\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds n^3\) |
$\blacksquare$