Integer to Power of Itself Falling is Factorial
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Theorem
Let $n \in \Z_{\ge 0}$ be a positive integer.
- $n^{\underline n} = n!$
where:
- $n^{\underline n}$ denotes the falling factorial
- $n!$ denotes the factorial.
Proof
\(\ds n^{\underline n}\) | \(=\) | \(\ds \dfrac {n!} {\paren {n - n}!}\) | Falling Factorial as Quotient of Factorials | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {n!} {0!}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds n!\) | Factorial of Zero |
$\blacksquare$