 Solution

Multiplication by the displayed Vandermonde matrix maps the coefficient vector of a polynomial of degree at most $n$ to its vector of values at the nodes. Part (a) proves that every vector of node values is attained uniquely, so this square linear map is bijective. Therefore $A_n$ is invertible. Equivalently, a kernel vector would define a polynomial of degree at most $n$ with $n+1$ distinct roots and would have to vanish.