Lagrange leading-coefficient identity (${\sum }_i{x}_i^k/{\prod }_{j\ne i}(x_i-x_j)={\delta }_{kn}$)

For $n+1$ distinct interpolation nodes and an integer $0\le k\le n$,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.