Theorem.  Let the column-vector of functions of,n

                                         T
                        a(n):=, [a[0](n)] , be defined

                                                          T
                    by the initial conditions:, a(0) = [1]

          and a(n+1)=A(n)a(n), where A(n) is the, 1,  by , 1,  matrix

                                    [[1/2]]

                                                                 T
              In other words, a(n)=A(n-1)A(n-2)...A(0) times, [1]

                                      Let

                        F(n, k) = binomial(n, k) a[0](n)

                   We have, for every non-negative integer n

                               -----
                                \
                                 )   F(n, k) = 1
                                /
                               -----
                                 k



                          Proof: We cleverly construct

                                   binomial(n, k) a[0](n) k
                    G(n, k) = -1/2 ------------------------
                                          n + 1 - k

                              with the motive that

                 F(n + 1, k) - F(n, k) = G(n, k + 1) - G(n, k)

                                    (check!)

          and the identity follows upon summing with respect to k. QED