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

\(\displaystyle n^{\underline n}\) \(=\) \(\displaystyle \dfrac {n!} {\paren {n - n}!}\) Falling Factorial as Quotient of Factorials
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {n!} {0!}\)
\(\displaystyle \) \(=\) \(\displaystyle n!\) Factorial of Zero

$\blacksquare$