Product of Number by its Falling Factorial

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Theorem

Let $x^{\underline n}$ denote the $n$th falling factorial power of $x$.

Then:

$x x^{\underline n} = x^{\underline {n + 1} } + n x^{\underline n}$


Proof

\(\ds x x^{\underline n}\) \(=\) \(\ds \paren {x - n + n} x^{\underline n}\)
\(\ds \) \(=\) \(\ds \paren {x - n} x^{\underline n} + n x^{\underline n}\)
\(\ds \) \(=\) \(\ds \paren {x - n} \prod_{j \mathop = 0}^{n - 1} \paren {x - j} + n x^{\underline n}\) Definition of Falling Factorial
\(\ds \) \(=\) \(\ds \prod_{j \mathop = 0}^n \paren {x - j} + n x^{\underline n}\) Definition of Continued Product
\(\ds \) \(=\) \(\ds x^{\underline {n + 1} } + n x^{\underline n}\) Definition of Falling Factorial

$\blacksquare$