Discussion
Let
\[ \begin{aligned} x^{(n)} &= x(x - h)(x - 2h) \cdots (x - (n - 2)h) \\ &\quad (x - (n - 1)h), \end{aligned} \]
where \(h\) is the step size. Then which of the following is correct ?
Answer: A
The correct pair is \(\Delta^{n} x^{(n)} = n!\,h^{n}\) and \(\Delta^{n+1} x^{(n)} = 0\).
The factorial polynomial behaves under \(\Delta\) (step \(h\)) the way \(x^{n}\) behaves under differentiation:
\[ \begin{aligned} \Delta x^{(n)} &= nh\, x^{(n-1)} \\ \Delta^{2} x^{(n)} &= n(n-1)h^{2}\, x^{(n-2)} \\ \Delta^{n} x^{(n)} &= n!\, h^{n} \end{aligned} \]
The \(n\)th difference is the constant \(n!\,h^{n}\), so one more difference gives 0. This matches the general rule that the \((n+1)\)th difference of any polynomial of degree \(n\) vanishes.
\(n!\) alone is the case \(h = 1\), and options claiming \(\Delta^{n+1} x^{(n)} \neq 0\) or \((n+1)!\) contradict that rule.
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