Sum of Lagrange polynomials: $\sum_{i=0}^{n}L_i(0)x_i^{n+1} = (-1)^{n}x_0\cdot\cdots\cdot x_n $. Given $\{x_0,...,x_n\}$ I am asked to show that $\sum_{i=0}^{n}L_i(0)x_i^{n+1} = (-1)^{n}x_0\cdot\cdots\cdot x_n $. 
I already showed that $\sum_{i=0}^{n}L_i(x)x_i^{j} = x^j$ for $j=1,...,n$ and that $\sum_{i=0}^{n}L_i(x)=1$ but I don't know how to use this to solve my problem. 
If I use the same method by which I solve the other problems then I am basically looking for a polynomial P of degree less than n such that $P(x_i) = x_i^{n+1}$ and 
$P(0) = (-1)^{n}x_0\cdot\cdots\cdot x_n$

Hint: $$P(x)=x^{n+1}-(x-x_0)(x-x_1)\cdot\cdot\cdot(x-x_n)$$ is a polynomial of degree $n$ or less.
