Definition The divided difference of function $f$ at points $x_0, x_1, ..., x_k$ is defined recursively as $$ f[x]=f(x), \qquad f[x_0,...,x_k] =\frac{f[x_1,...,x_k]-f[x_0,...,x_{k-1}]}{x_k-x_0}, \quad k \ge 1 $$
Prove that if $f(x) = x^n$ for $n\in\mathbb{N}$, then $f[x_0,...,x_{n-1}] = \sum_{i=0}^{n-1}x_i$.
I initially thought about induction but the problem is that change in $n$ causes change in $f$ and I was unable to derive a useful formula for dependency between $a^{n-1}-b^{n-1}$ and $a^n-b^n$.
Example If $f(x) = x^3$ then $$f[a,b,c]=\frac{f[b,c]-f[a,b]}{c-a}=\frac{\frac{f[c]-f[b]}{c-b}-\frac{f[b]-f[a]}{b-a}}{c-a}=\frac{\frac{c^3-b^3}{c-b}-\frac{a^3-b^3}{a-b}}{c-a}=a+b+c$$