and based on nodes $x_0,$dots,$x_n,$ show that for any $f,$

${\displaystyle \underset{i=0}{\overset{n}{}}}f(x_i)L_i(x)={\displaystyle \underset{i=0}{\overset{n}{}}}f[x_0,$dots,$x_i]pro{d}_j=0^{i-1}(x-x_j)$