Rising Factorial in terms of Falling Factorial of Negative

Theorem

$x^{\overline k} = \paren {-1}^k \paren {-x}^{\underline k}$

where:

$x^{\underline k}$ is the $k$ to the $x$ falling

$x^{\overline k}$ is the $k$ to the $x$ rising.

$\ds \paren {-x}^{\underline k}$

$=$

$\ds \prod_{j \mathop = 0}^{k - 1} \paren {-x - j}$

$\ds $

$=$

$\ds \prod_{j \mathop = 0}^{k - 1} \paren {-1} \paren {x + j}$

$\ds $

$=$

$\ds \prod_{j \mathop = 0}^{k - 1} \paren {-1} \prod_{j \mathop = 0}^{k - 1} \paren {x + j}$

$\ds $

$=$

$\ds \paren {-1}^k \prod_{j \mathop = 0}^{k - 1} \paren {x + j}$

$\ds $

$=$

$\ds \paren {-1}^k x^{\overline k}$

$\ds \leadsto \ \ $

$\ds \paren {-1}^k \paren {-x}^{\underline k}$

$=$

$\ds \paren {-1}^{2 k} x^{\overline k}$

multiplying both sides by $\paren {-1}^k$

$\ds $

$=$

$\ds x^{\overline k}$

$-1$ to an even power is $1$

$\blacksquare$