= Lagrange leading-coefficient identity
{c}
{title2=$\sum_i x_i^k/\prod_{j\ne i}(x_i-x_j)=\delta_{kn}$}
For $n+1$ distinct <interpolation nodes> and an integer $0\leq k\leq n$,
$$
\sum_{i=0}^n\frac{x_i^k}{\prod_{j\ne i}(x_i-x_j)}
=\begin{cases}0,&k<n,\\1,&k=n.\end{cases}
$$
To prove this, the <Lagrange interpolation polynomial> for the data $x_i^k$ is exactly $x^k$, by uniqueness of <polynomial interpolation>. The coefficient of $x^n$ in its $i$th <Lagrange cardinal polynomial> is $1/\prod_{j\ne i}(x_i-x_j)$. Comparing the coefficients of $x^n$ gives the identity. At $k=0$, values of the constant <polynomial> $1$ are intended even at a zero node. Negative exponents are outside the assertion.
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