A Combinatorial Proof of Vandermonde's Determinant

that is as easy as playing cards. Let Vn denote the Vandermonde matrix with (i, j)th entry vi j = x j i (0 ≤ i, j ≤ n). Since the determinant of Vn is a polynomial in x0, x1, . . . , xn , it suffices to prove the identity for positive integers x0, x1, . . . , xn with x0 ≤ x1 ≤ · · · ≤ xn . We define a Vandermonde card to possess a suit and a value, where a card of Suit i has a value from the set {1, . . . , xi }. (In our examples, we will let Suits 0, 1, 2, 3, and 4 be represented by suits ,♣,♦,♥, and ♠, respectively.) Hence there are x0 + x1 + · · · + xn different Vandermonde cards, but we have at our disposal an unlimited supply of each. First we do some card counting.