Prove $f[x_0,...,x_{n-1}] = \sum_{i=0}^{n-1}x_i$ if $f(x) = x^n$ for $n\in\mathbb{N}$ 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 $$f[a,b,c]=\frac{f[b,c]-f[a,b]}{c-a}=\frac{\frac{f[c]-f[b]}{c-b}-\frac{f[b]-f[a]}{b-a}}{c-a}=\frac{\frac{c^3-b^3}{c-b}-\frac{a^3-b^3}{a-b}}{c-a}=a+b+c$$
 A: Here's a sketch of one solution.First see that $f[x_0,\cdots,x_{n-1}]$ is a polynomial of degree $1$. Use the fact that they are symmetric to see that we must have
$$f[x_0,\cdots,x_{n-1}]=k\sum_{i=0}^{n-1} x_i$$
for some $k$. Now use their mean value theorem to conclude.
A: We consider Newton's form of the interpolating polynomial. Let $f(x) = x^n$ and let $\{x_j\}_{j=0}^n$ denote $n+1$ distinct nodes. Then 
$$x^n = f[x_0] + f[x_0,x_1](x-x_0) + \dots + f[x_0,\dotsc,x_{n-1}] \prod_{i=0}^{n-2}(x-x_i) + f[x_0,\dotsc,x_n] \prod_{i=0}^{n-1}(x-x_i).$$
We will exploit the fact that the coefficient of $x^{n-1}$ is $0$. We begin by studying the structure of the different terms. The last term is \begin{align}
& f[x_0,\dotsc,x_n] \prod_{i=0}^{n-1}(x-x_i) \\
& = f[x_0,\dotsc,x_n] x^n - f[x_0,\dotsc,x_n] (x_0 + x_1 + \dots + x_{n-1}) x^{n-1} + O(x^{n-2}),
\end{align}
where $O(x^{n-2})$ means "terms with $x$-degree $\leq n-2$".
It is possible to conclude that $$f[x_0,\dotsc,x_n] = 1,$$ because all other terms have order at most $n-1$ and the coefficient of $x^n$ is $1$ (alternatively, see this post). As for the second to last term we have 
$$ f[x_0,\dotsc,x_{n-1}] \prod_{i=0}^{n-2}(x-x_i) = f[x_0,\dotsc,x_{n-1}] x^{n-1} + O(x^{n-2}).$$
Since all other terms have degree at most $n-2$ we conclude that 
$$ f[x_0,\dotsc,x_{n-1}] = x_0 + x_1 + \dots + x_{n-1}.$$
It is worth stressing our use of fact that two polynomials are identical if and only their coefficients are identical. This completes the analysis.
A: The Liebniz rule states that 
$$
(fg)[x_0,\dots,x_n] = f[x_0]g[x_0,\dots,x_n] + f[x_0,x_1]g[x_1,\dots,x_n] + \dots + f[x_0,\dots,x_n]g[x_n].
$$
Proceed by induction on $n$. For convenience, let $f_n(x) := x^n$. Notice that the base case is trivial. 
Assume that the result holds for some positive integer $k$. It can be shown that 
\begin{equation} 
f_1[x_0,\dots,x_j] = 
\begin{cases}
1, & j=1; \\
0, & j>1.
\end{cases} \tag{1} \label{ddi}
\end{equation}
Thus, 
\begin{align}
f_{k+1}[x_0,\dots,x_k] 
&= \left(f_1 f_k\right)[x_0,\dots,x_k] \\
&= f_1[x_0]f_k[x_0,\dots,x_k] + 
f_1[x_0, x_1]f_k[x_1,\dots,x_k] + \cdots + f_1[x_0,\dots,x_k]f_k[x_k]     \\
&= x_0 f_k[x_0,\dots,x_k] + f_1[x_0,x_1]f_k[x_1,\dots,x_k]    \tag{by \eqref{ddi}} \\
&= x_0 f_k[x_0,\dots,x_k] + \sum_{j=1}^{k} x_j.                \tag{IH}
\end{align}
By the mean value theorem for divided differences, $\exists \xi \in \left[ \min_j\{x_j\},\max_j\{x_j\} \right]$ such that 
$$
f_k[x_0,\dots,x_k] = \frac{f_k^{(k)}(\xi)}{k!}.
$$
But $f_k^{(k)}(x) = k!$, so $f_k[x_0,\dots,x_k]=1$. Thus, 
$$
f_{k+1}[x_0,\dots,x_k] = \sum_{j=0}^k x_j
$$
and the result follows by induction.
