$\nabla^nf(x_i)=\sum\limits_{k=0}^{n}(-1)^k\frac{n!}{k!(n-k)!}f(x_i-kh).$

Question

Can anyone help me prove this $$\nabla^nf(x_i)=\sum\limits_{k=0}^{n}(-1)^k\frac{n!}{k!(n-k)!}f(x_i-kh).$$ where $\nabla f(x_i)=f(x_i)-f(x_i-h).$
I tried using induction but I'm not getting anywhere.

Answer 1

$\nabla^{n+1}f(x_i)=\sum\limits_{k=0}^{n}(-1)^k\frac{n!}{k!(n-k)!}\nabla f(x_i-kh)$

$=\sum\limits_{k=0}^{n}(-1)^k\frac{n!}{k!(n-k)!}(f(x_i-kh)-f(x_i-(k+1)h))$

$=\sum\limits_{k=0}^{n}(-1)^k\frac{n!}{k!(n-k)!}f(x_i-kh)-\sum\limits_{k=1}^{n+1}(-1)^{k-1}\frac{n!}{(k-1)!(n-k+1)!}f(x_i-kh)$

$=\sum\limits_{k=0}^{n+1}(-1)^k\frac{(n+1)!}{k!(n+1-k)!}\nabla f(x_i-kh)$

because of $\enspace \frac{n!}{k!(n-k)!}+\frac{n!}{(k-1)!(n-k+1)!}=\frac{(n+1)!}{k!(n+1-k)!} \enspace$ and $\frac{1}{(-1)!}=0$ .