Number to Power of One Falling is Itself

Theorem

Let $x \in \R$ be a real number.

$x^{\underline 1} = x$

where $x^{\underline 1}$ denotes the falling factorial.

Proof

\(\ds x^{\underline 1}\)

\(=\)

\(\ds \dfrac {\Gamma \left({x + 1}\right)} {\Gamma \left({x - 1 + 1}\right)}\)

Falling Factorial as Quotient of Factorials

\(\ds \)

\(=\)

\(\ds \dfrac {\Gamma \left({x + 1}\right)} {\Gamma \left({x}\right)}\)

\(\ds \)

\(=\)

\(\ds x\)

Gamma Difference Equation

$\blacksquare$