<s>[INST] Proving that $\Delta x^{(n)} = n x^{(n-1)}$ Define $\Delta f(x) = f(x+1) - f(x)$ (the difference operator). Define $x^{(n)} = x(x-1) \dots (x-n+1)$ (the falling factorial function). There's a rather simple theorem which shows that $\Delta x^{(n)} = n x^{(n-1)}$:
$\Delta x^{(n)} = (x+1)^{(n)} - x^{(n)}$
$= (x+1)(x+1-1)(x+1-2)\dots (x+1 - n+1) - x(x-1) \dots (x-n+2)(x-n+1)$
$= (x+1)(x)(x-1) \dots (x-n+2) - x(x-1) \dots (x-n+2)(x-n+1)$
$= [(x+1) - (x-n+1)][(x)(x-1) \dots (x-n+2)]$
$= n [(x)(x-1) \dots (x-n+2)]$
$= n x^{(n-1)}$
My (probably silly) question is about the step from the second line to the third line (the rest of the proof is straightforward). Shouldn't the third line contain the expression $(x+1)(x)(x-1) \dots (x-n)$, instead of $(x+1)(x)(x-1) \dots (x-n+2)$?
 [/INST] That first term is the product of $n$ factors of the form $x+1-k$, where $k$ starts at $0$ and increases by $1$ from one factor to the next. The highest value of $k$ must therefore by $n-1$, and the last factor is therefore $x+1-(n-1)=x-n+2$.
If you went all the way to $x-n=(x+1)-(n+1)$, you’d have $k$ running from $0$ to $n+1$, for a total of $n+2$ factors.
</s>