Show using the Newton representation of interpolation polynomial that $y[x_0,x_1,\ldots ,x_n]=0$ if $y_i=p(x_i)$, $i=0, \ldots , n$ for a polynomial $p$ of degree $n-1$.
The Newton representation of interpolation polynomial using the divided differences is \begin{align*}p(x)&=\sum_{i=0}^ny[x_0,x_1,\ldots ,x_i](x-x_0)\cdots (x-x_{i-1})\end{align*}
We have that $$p(x)= y[x_0] + (x - x_0)y[x_0, x_1] + \ldots + \prod_{j = 0}^{n - 1} (x - x_j) y[x_0, \ldots, x_n]$$ Do we take $x=x_i$ for $0\leq i\leq n$ to get desired result?
Let $q$ be the unique polynomial of degree at most $n$ such that $q(x_i) = y(x_i)$ for $i = 0, 1, \dots, n$. Then by Newton's representation, we have $$q(x) = \sum_{i = 0}^{n}y[x_0, \dots, x_i](x - x_0)\dots(x - x_{i - 1}).$$ By uniqueness, $q = p$. Therefore the coefficient of $x^n$ in $q(x)$ is $0$. This means exactly that $$y[x_0, \dots, x_n] = 0.$$