Proof

Let $x\in \mathbb C$ is a complex number and let $n\ge 0$ be non-negative integer.
Case $n\ge 1$:
- Then, by definition of the difference operator, we get for the falling factorial power $$\begin{align}\Delta x^{\underline {n}}&=(x+1)^{\underline {n}}-x^{\underline {n}}\nonumber\\ &=(x+1)[x(x-1)\cdots(x-n+2)]-\nonumber\\ &\quad-[x(x-1)\cdots(x-n+2)(x-n+1)]\nonumber\\ &=[(x+1)-(x-n+1)]\cdot[x(x-1)\cdots(x-n+2)]\nonumber\\ &=nx^{\underline {n-1}}.\nonumber \end{align}$$
Case $n=0$, $x\neq 1$:
- By definition of the difference operator and the falling factorial power $$\Delta x^{\underline {0}}=(x+1)^{\underline {0}}-x^{\underline {0}}=1-1=0=0\cdot x^{\underline {-1}}.$$
Case $-n<0$, $x\not\in\{1,2,\ldots,n,n+1\}$
- By definition of the difference operator and the falling factorial power. $$\begin{align}\Delta x^{\underline {-n}}&=\frac{1}{{x+1}^{\underline n}}-\frac{1}{x^\underline n}\nonumber\\ &=\frac{1}{(x+1)[x(x-1)\cdots(x-n-2)]}-\frac{1}{[x(x-1)\cdots(x-n-2)](x-n-1)}\nonumber\\ &=\frac{x-n-1}{(x+1)[x(x-1)\cdots(x-n-2)](x-n-1)}-\frac{x+1}{(x+1)[x(x-1)\cdots(x-n-2)](x-n-1)}\nonumber\\ &=-n\cdot \frac{1}{x^{\underline{n+1}}}\nonumber\\ &=-n\cdot x^{\underline{-n-1}}\nonumber \end{align}$$
It follows that $\Delta x^{\underline{n}}=nx^{\underline{n-1}}$ for all integers $n\in\mathbb Z.$
∎

Thank you to the contributors under CC BY-SA 4.0!

Miller, Kenneth S.: "An Introduction to the Calculus of Finite Differences And Difference Equations", Dover Publications, Inc, 1960