Q:

Application of Lagrange Polynomials

Let $L_i(x)$ be the Lagrange polynomial for pairwise different nodal point $x_0,x_1,\cdots,x_n$, and let $c_i=L_i(0)$. 
I proved that
\begin{equation}
(1) \sum_{i=0}^{n} L_i(x)=1\\
(2) \sum_{i=0}^{n} L_i(x){x_i}^k=x^k \space\space\space\space \text{for} \space  k=1,2,\cdots,n
\end{equation}
I cannot prove $k=n+1$ case 
\begin{equation}
\sum_{i=0}^{n} L_i(0){x_i}^{n+1}=(-1)^n x_0x_1\cdots x_n
\end{equation}
I could represent above left term as 
\begin{equation}
(-1)^nx_0x_1\cdots x_n(\frac{{x_0}^n}{(x_0-x_1)\cdots(x_0-x_n)}+\frac{{x_1}^n}{(x_1-x_0)(x_1-x_2)\cdots(x_1-x_n)}+\cdots +\frac {{x_n}^n}{(x_n-x_0)\cdots(x_n-x_{n-1})})
\end{equation}
Thus, it suffices to prove that 
\begin{equation}
\frac{{x_0}^n}{(x_0-x_1)\cdots(x_0-x_n)}+\frac{{x_1}^n}{(x_1-x_0)(x_1-x_2)\cdots(x_1-x_n)}+\cdots +\frac {{x_n}^n}{(x_n-x_0)\cdots(x_n-x_{n-1})}=1
\end{equation}
Any help would be appreciated!!! 

A:

The polynomial $p(x)=x^{n+1}-(x-x_0)\dots(x-x_n)$ is of order $n$, so $p$ is its own interpolant (at the nodes $x_0,\dots,x_n$).  The Lagrange form is:
$$
 p(x)=x^{n+1}-(x-x_0)\dots(x-x_n)=\sum_{i=0}^nL_i(x)p(x_i)=\sum_{i=0}^nL_i(x)x_i^{n+1}. 
$$
It is easy to observe by computing the values of the left hand side at $x_0,\dots,x_{n+1}$. Now it is enough to put $x=0$ here.