Divided differences of $x^n$

Denote $f(x)=x^n$ ($n$ is natural), I am trying to prove that $f[x_0,x_1,\ldots,x_n]=1$, $\{x_i\}$ are distinct $n+1$ real numbers.

I tried doing this by finding an interpolation of the function and by using Newton's form but what I did didn't lead to the solution.

I also tried using the definition as given in Wikipidia but that didn't help me either.

any ideas on how to prove this ?

[This question is taken from the book Numerical analysis by David Ronald Kincaid‏, Elliott Ward Cheney‏]

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Since $f[x_0,\dots,x_n]$ is constant, the statement follows from equality $$\lim_{(x_0,\dots,x_n)\to(\xi,\dots,\xi)} f[x_0,\dots,x_n] = \frac{f^{(n)}(\xi)}{n!}$$ in Wikipedia. – Pierre-Yves Gaillard Dec 31 '11 at 14:54
You should probably turn this into an answer, Pierre. This is the key to OP's problem. – Patrick Da Silva Dec 31 '11 at 15:03
Dear @Patrick: Thanks!!! [I saw your comment by chance. It's safer (if you think of it) to use the @sign to ping someone.] – Pierre-Yves Gaillard Dec 31 '11 at 15:27
I know how to ping : sorry I didn't do it. – Patrick Da Silva Dec 31 '11 at 15:32
Dear @Patrick: Please, don't be sorry. The main point of my comment was to thank you. – Pierre-Yves Gaillard Dec 31 '11 at 15:53

Let $x_0,x_1,\dots$ be distinct real numbers, let $f$ be a polynomial function on $\mathbb R$, and define $f[x_0,\dots,x_j]$ for $j=0,1,\dots$ recursively by $$f[x_0]:=f(x_0),$$ $$f[x_0,\dots,x_j]:=\frac{f[x_1,\dots,x_j]-f[x_0,\dots,x_{j-1}]}{x_j-x_0}\quad,\quad j\ge1.$$
It is straightforward to check that $f[x_0,\dots,x_j]$ is a homogeneous polynomial of degree $d-j$ in $x_0,\dots,x_j$ if $f$ is homogeneous of degree $d$.
In your case, $f[x_0,\dots,x_n]$ is constant, and the statement follows from equality $$\lim_{(x_0,\dots,x_n)\to(\xi,\dots,\xi)} f[x_0,\dots,x_n] = \frac{f^{(n)}(\xi)}{n!}$$ in Wikipedia.