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 
$P(0) = (-1)^{n}x_0\cdot\cdots\cdot x_n$
 A: 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.
A: It can be proven with determinant form of Lagrange polynomial that interpolates $(x_0;y_0)$, $\dots$, $(x_n;y_n)$
$$
P(x) = (-1)
\frac{
    \det
    \begin{pmatrix}
        0       & y_0       & y_1       & \cdots & y_n       \\
        x^n     & x_0^n     & x_1^n     & \cdots & x_n^n     \\
        x^{n-1} & x_0^{n-1} & x_1^{n-1} & \cdots & x_n^{n-1} \\
        \cdots  & \cdots    & \cdots    & \cdots & \cdots    \\
        1       & 1         & 1         & \cdots & 1         \\
    \end{pmatrix}
}{
    \det
    \begin{pmatrix}
        x_0^n     & x_1^n     & \cdots & x_n^n     \\
        x_0^{n-1} & x_1^{n-1} & \cdots & x_n^{n-1} \\
        \cdots    & \cdots    & \cdots & \cdots    \\
        1         & 1         & \cdots & 1         \\
    \end{pmatrix}
},
$$
presented in "Beginner's guide to mapping simplexes affinely", section "Lagrange interpolation".
Obviously, setting all $y_i = x_i^{n+1}$ and $x = 0$ we'll get the expression you need, because
$$
    P(x) = \sum_{i=0}^n\, y_i L_i(x) = \sum_{i=0}^n x_i^{n+1} L_i(x).
$$
So I just plug them in
$$
P(0) = (-1)
\frac{
    \det
    \begin{pmatrix}
        0       & x_0^{n+1} & x_1^{n+1} & \cdots & x_n^{n+1} \\
        0       & x_0^n     & x_1^n     & \cdots & x_n^n     \\
        0       & x_0^{n-1} & x_1^{n-1} & \cdots & x_n^{n-1} \\
        \cdots  & \cdots    & \cdots    & \cdots & \cdots    \\
        1       & 1         & 1         & \cdots & 1         \\
    \end{pmatrix}
}{
    \det
    \begin{pmatrix}
        x_0^n     & x_1^n     & \cdots & x_n^n     \\
        x_0^{n-1} & x_1^{n-1} & \cdots & x_n^{n-1} \\
        \cdots    & \cdots    & \cdots & \cdots    \\
        1         & 1         & \cdots & 1         \\
    \end{pmatrix}
}
$$
Now consider Laplace expansion along the first column
$$
P(0) = (-1)^n
\frac{
    \det
    \begin{pmatrix}
        x_0^{n+1} & x_1^{n+1} & \cdots & x_n^{n+1} \\
        x_0^n     & x_1^n     & \cdots & x_n^n     \\
        x_0^{n-1} & x_1^{n-1} & \cdots & x_n^{n-1} \\
        \cdots    & \cdots    & \cdots & \cdots    \\
        x_0       & x_1       & \cdots & x_n       \\
    \end{pmatrix}
}{
    \det
    \begin{pmatrix}
        x_0^n     & x_1^n     & \cdots & x_n^n     \\
        x_0^{n-1} & x_1^{n-1} & \cdots & x_n^{n-1} \\
        \cdots    & \cdots    & \cdots & \cdots    \\
        1         & 1         & \cdots & 1         \\
    \end{pmatrix}
}
$$
and use properties of determinant
$$
P(0) = (-1)^n x_0 x_1 \dots x_n
\frac{
    \det
    \begin{pmatrix}
        x_0^n     & x_1^n     & \cdots & x_n^n     \\
        x_0^{n-1} & x_1^{n-1} & \cdots & x_n^{n-1} \\
        \cdots    & \cdots    & \cdots & \cdots    \\
        1         & 1         & \cdots & 1         \\
    \end{pmatrix}
}{
    \det
    \begin{pmatrix}
        x_0^n     & x_1^n     & \cdots & x_n^n     \\
        x_0^{n-1} & x_1^{n-1} & \cdots & x_n^{n-1} \\
        \cdots    & \cdots    & \cdots & \cdots    \\
        1         & 1         & \cdots & 1         \\
    \end{pmatrix}
}.
$$
The latter finishes the proof
$$
   P(0) = (-1)^n x_0 x_1 \dots x_n
$$
For more practical examples you may want to check "Workbook on mapping simplexes affinely", section "Lagrange interpolation".
