Let be $N_{n+1}(x)=\prod_{i=0}^n(x-x_i)$. Now I have to prove that
$$||N_{n+1}(x)||_{\infty,[-5,5]}\leq n!\frac{h^{n+1}}{4},\qquad h:=\frac{5-(-5)}{n}=\frac{10}{n}.$$
I've started with
$$\|N_{n+1}(x)\|_{\infty,[-5,5]}=\max_{x\in[-5,5]}\left|\prod_{i=0}^n(x-x_i) \right|.$$ Could anyone help me, please?