Let A be a contraction on a Hilbert space H . The defect index d A of A is, by definition, the dimension of the closure of the range of I - A ∗ A . We prove that (1) d A n ⩽ nd A for all n ⩾ 0 , (2) if, in addition, A n converges to 0 in the strong operator topology and d A = 1 , then d A n = n for all finite n , 0 ⩽ n ⩽ dim H , and (3) d A = d A ∗ implies d A n = d A n ∗ for all n ⩾ 0 . The norm-one index k A of A is defined as sup { n ⩾ 0 : ‖ A n ‖ = 1 } . When dim H = m < ∞ , a lower bound for k A was obtained before: k A ⩾ ( m / d A ) - 1 . We show that the equality holds if and only if either A is unitary or the eigenvalues of A are all in the open unit disc, d A divides m and d A n = nd A for all n , 1 ⩽ n ⩽ m / d A . We also consider the defect index of f ( A ) for a finite Blaschke product f and show that d f ( A ) = d A n , where n is the number of zeros of f .