Proof

Let f(X) R[X] be a polynomial of degree at most n - 1. Write f(X) = f₀ + f₁X + ··· + f_n-1X^n-1 and substitute the given values. This transforms the problem into that of solving the system of linear equations:

f₀ + f₁ · x₁ + ··· + f_n-1 · x₁^n-1 = a₁

f₀ + f₁ · x₂ + ··· + f_n-1 · x₂^n-1 = a₂

·                      · ·                      · ·                      ·

f₀ + f₁ · x_n + ··· + f_n-1 · x_n^n-1 = a_n.

This system can be rewritten in matrix form as M f = a, where f is the `vector' (f₀, ..., f_n-1)^T, a the vector (a₁, ..., a_n)^T, and M the matrix M = (x_i^{j - 1})_{i = 1,..., n, j = 1,..., n}. The polynomial (equivalently, the vector) f are the n unknowns.

Now M is a so-called Vandermonde matrix. It has the special property that the determinant is given by the formula

det M = _i<j (x_j - x_i). Since the x_i are chosen to be distinct, the determinant is nonzero. Since R is a field, the inverse of the determinant exists and M is invertible. This means that the system of linear equations has exactly one solution.