[proofplan] Uniqueness follows from the fact that a nonzero polynomial of degree at most $n$ has at most $n$ real roots: if two interpolants agreed at $n+1$ points, their difference would be a polynomial of degree $\le n$ with $n+1$ roots, hence zero. Existence is given by the Lagrange construction: define cardinal polynomials $\ell_i$ that are $1$ at $x_i$ and $0$ at every other node, then take $p = \sum f_i\,\ell_i$. [/proofplan] [step:Prove uniqueness via the root-counting argument] Suppose $p, q \in P_n[x]$ both satisfy $p(x_i) = q(x_i) = f_i$ for $i = 0, \dots, n$. Then $r := p - q \in P_n[x]$ vanishes at $n+1$ distinct points $x_0, \dots, x_n$. Since a nonzero polynomial of degree at most $n$ has at most $n$ real roots, $r$ must be the zero polynomial. Therefore $p = q$. [/step] [step:Construct the interpolant via Lagrange cardinal polynomials] Define the Lagrange cardinal polynomials \begin{align*} \ell_i(x) &= \prod_{\substack{j=0 \\ j \neq i}}^{n} \frac{x - x_j}{x_i - x_j}, \quad i = 0, \dots, n. \end{align*} Each $\ell_i$ is a polynomial of degree $n$. It satisfies $\ell_i(x_j) = \delta_{ij}$: it equals $1$ at $x_i$ (since each factor $(x_i - x_j)/(x_i - x_j) = 1$) and vanishes at every other node $x_j$ with $j \neq i$ (since the factor $(x_j - x_j)/(x_i - x_j) = 0$). Define \begin{align*} p(x) &= \sum_{i=0}^n f_i\,\ell_i(x). \end{align*} This is a polynomial of degree at most $n$, and for each $j = 0, \dots, n$: \begin{align*} p(x_j) &= \sum_{i=0}^n f_i\,\delta_{ij} = f_j. \end{align*} [/step]