Tips to demonstrate for that $x^{\underline{n+p}}=n^{\underline{n}}\:\left(x-n\right)^{\underline{p}}$

Question

I have to prove for $x\:\in \:\mathbb{R}\:$ and $\:n,\:p\:\in\:\mathbb{Z}$, that:

$x^{\underline{n+p}}=n^{\underline{n}}\:\left(x-n\right)^{\underline{p}}$

Without resolving with Capital pi notation, how you would resolve it?

My idea is to apply induction only on n, and then on p but i am not sure with i have to analyse the following cases when:

(I) $n\:=\:0,\:p\:=0$

(II) $n\:=\:0\:and\:p\:\in \:\mathbb{Z}^+$

(III) $p\:=\:0\:and\:n\:\in \:\mathbb{Z}^+$

(IV) $n\:\in \:\mathbb{Z}^+\:and\:p\:\in \:\:\mathbb{Z}^+$

(V) $n\:=\:0\:and\:p\:\in \:\mathbb{Z}^-$

(VI) $p\:=\:0\:and\:n\:\in \:\mathbb{Z}^-$

(VII) $n\:\in \:\mathbb{Z}^-\:and\:p\:\in \:\:\mathbb{Z}^+$

Answer 1

This follows directly from the definition: $$ m^{\underline{n}}=\frac{m!}{(m-n)!}=\frac{\Gamma(m+1)}{\Gamma(m-n+1)}. $$ We write \begin{align} x^{\underline{n+p}} &=\frac{\Gamma(x+1)}{\Gamma(x-n-p+1)}\\ &=\frac{\Gamma(x+1)}{\Gamma(x-n-p+1)}\cdot\underbrace{\frac{\Gamma(x-n+1)}{\Gamma(x-n+1)}}_{=1}\\ &=\frac{\Gamma(x+1)}{\Gamma(x-n+1)}\cdot\frac{\Gamma(x-n+1)}{\Gamma(x-n-p+1)}\\ &=x^{\underline{n}}(x-n)^{\underline{p}}. \end{align}

Answer 2

It is a PITA to prove all of these cases. We can make our work simpler by proving some general facts out iterated products.

To be clear, given $f:\mathbb Z\to \mathbb C$, and $a,b\in \Bbb Z$, let us define $$ \prod_{i=a}^b f(i)= \begin{cases} f(a)\cdot f(a+1)\cdots f(b) & b > a-1 \\ 1 & b = a-1 \\ f(a-1)^{-1}f(a-2)^{-1}\cdots f(b+1)^{-1} & b <a-1 \end{cases} $$ The key property to prove is that for any $a,b,c\in \mathbb Z$, we have $$ \left(\prod_{i=a}^{b-1}f(i)\right)\left(\prod_{i=b}^{c-1}f(i)\right)=\prod_{i=a}^{c-1} f(i) $$ This has to be proven in $9$ cases, based on whether we have $a<b,a=b,$ or $a>b$, and the same considerations for $b$ and $c$. All of the proofs are tedious at worst.

Once we have that general fact under our belt, your problem is handled easily: $$ \begin{align} x^{\underline{n+p}} &=\prod_{i=0}^{n+p-1} (x-i)=\left(\prod_{i=0}^{n-1}(x-i)\right)\left(\prod_{i=n}^{n+p-1}(x-i)\right)=x^{\underline n}\left(\prod_{i=0}^{p-1}(x-n-i)\right)=x^{\underline n}\cdot (x-n)^{\underline p} \end{align} $$ In the second equality, we use the general property of $\prod$, and the the third equality, we re-index the second sum.