= Solution
{c}
The <monomial> $x^k$ lies in $\Pi_n$ and already fits the data. Uniqueness of <polynomial interpolation> therefore gives
$$
\boxed{p_n(x)=x^k.}
$$
Apply this result with a nonnegative integer $m\leq n$. Its <Lagrange interpolation> representation is the <polynomial identity>
$$
x^m=\sum_{i=0}^n x_i^m
\frac{\prod_{j\ne i}(x-x_j)}{\prod_{j\ne i}(x_i-x_j)}.
$$
The numerator in each summand is a <monic polynomial> of degree $n$, so the coefficient of $x^n$ on the right is exactly $\sum_i x_i^m/\prod_{j\ne i}(x_i-x_j)$. The coefficient on the left is zero for $m<n$ and one for $m=n$. Comparing coefficients proves the <Lagrange leading-coefficient identity>
$$
\boxed{\sum_{i=0}^n\frac{x_i^m}{\prod_{j\ne i}(x_i-x_j)}
=\begin{cases}0,&0\leq m<n,\\1,&m=n.\end{cases}}
$$
The intended exponent range is $m\geq0$: arbitrary negative exponents would not give this <polynomial identity>. For $m=0$, $x_i^0$ denotes the value of the constant <polynomial> $1$, including at a zero <interpolation node>.
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