Number to Power of Zero Falling is One

Theorem

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

$x^{\underline 0} = 1$

where $x^{\underline 0}$ denotes the falling factorial.

Proof

\(\ds x^{\underline 0}\)

\(=\)

\(\ds \prod_{j \mathop = 0}^{-1} \paren {x - j}\)

Definition of Falling Factorial

\(\ds \)

\(=\)

\(\ds 1\)

Product is Vacuous

$\blacksquare$