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Let $f(x)=x^n$ for $n \in \mathbb{N}$. Prove that $f[x_0,x_1,\ldots,x_n]=1$, where $\{x_i\}$ 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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As kindly suggested by Patrick Da Silva, I'm turning my comment into an answer.

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.

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A more simple approach proposed by the op @Belgi themselves is: By the mean value theorem for divided differences we have $$ f[x_0, \ldots, x_n] = \frac{f^{(n)}(\xi)}{n!} \qquad \text{for } \xi \in \big(\min(x_0, \ldots, x_n), \max(x_0, \ldots, x_n) \big). $$ Now, for $f(x) := x^n$ we have $f^{(n)}(\xi) = n (n - 1) \ldots \xi^0 = n!$, which is independent of $\xi$. Therefore, $$ f[x_0, \ldots, x_n] = \frac{f^{(n)}(\xi)}{n!} = \frac{n!}{n!} = 1. $$

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