Integer to Power of Itself Less One Falling is Factorial

Theorem

Let $n \in \Z_{\ge 0}$ be a positive integer.

$n^{\underline {n - 1} } = n!$

where:

$n^{\underline {n - 1} }$ denotes the falling factorial

$n!$ denotes the factorial.

Proof

\(\ds n^{\underline {n - 1} }\)

\(=\)

\(\ds \dfrac {n!} {\left({n - \left({n - 1}\right)}\right)!}\)

Falling Factorial as Quotient of Factorials

\(\ds \)

\(=\)

\(\ds \dfrac {n!} {1!}\)

\(\ds \)

\(=\)

\(\ds n!\)

Factorial of $1$

$\blacksquare$