Let us define the binomial coefficients, C(n,k), to be the coefficients
in the formal power series

         n   ---         k
    (1+x)  = >   C(n,k) x                                    [1]
             ---
              k

where C(n,k) = 0 for any k < 0.  Using the Taylor series for (1+x)^n,
we get for k >= 0,

             n(n-1)(n-2)(n-3)...(n-k+1)
    C(n,k) = --------------------------                      [2]
                         k!

When n is a non-negative integer, this becomes

                n!
    C(n,k) = --------                                       [3a]
             (n-k)!k!

and for -n a negative integer,

                  k
    C(-n,k) = (-1)  C(n+k-1,k)                              [3b]

Note that from the symmetry in [3a], we have that for non-negative n,

    C(n,k) = C(n,n-k)                                        [4]

Letting x = 1 and -1 in [1], we get for non-negative, integral n,

    ---           n
    >   C(n,k) = 2                                          [5a]
    ---
     k

and

    ---     k          n
    >   (-1) C(n,k) = 0                                     [5b]
    ---
     k

where 0^0 = 1.

Taking the product of [1] for two different exponents, we get

         n      m   --- ---                  k
    (1+x)  (1+x)  = >   >   C(n,k-j) C(m,j) x               [6a]
                    --- ---
                     k   j

and

         n+m   ---           k
    (1+x)    = >   C(n+m,k) x                               [6b]
               ---
                k

Equating the binomial coefficients in [6a] and [6b], we get the
identity

    ---
    >   C(n,k-j) C(m,j) = C(n+m,k)                           [7]
    ---
     j

Applying [3b], [4], and [7], we get

     n
    ---
    >   C(n-j,k) C(j,m)
    ---
    j=0

       n
      ---
    = >   C(n-j,n-j-k) C(j,j-m)
      ---
      j=0

       n
      ---     n-j-k                  j-m
    = >   (-1)     C(-k-1,n-j-k) (-1)   C(-m-1,j-m)
      ---
      j=0

      ---     n-k-m
    = >   (-1)     C(-k-1,n-j-k) C(-m-1,j-m)
      ---
       j

          n-k-m
    = (-1)     C(-m-k-2,n-k-m)


    = C(n+1,n-k-m)


    = C(n+1,k+m+1)                                           [8]

Note the following:

       1         1
    ------ - --------
    C(n,k)   C(n+1,k)

           1     n+1        1     n-k+1
    = ---------- --- - ---------- -----
      C(n+1,k+1) k+1   C(n+1,k+1)  k+1

       k       1
    = --- ----------                                         [9]
      k+1 C(n+1,k+1)

If we iterate [9] a few times, the following emerges:

    ---     j  C(m,j)     k       1
    >   (-1)  -------- = --- ----------                     [10]
    ---       C(n+j,k)   k+m C(n+m,k+m)
     j

We can prove [10] by induction.  It is obviously true for m = 0:

       1     k    1
    ------ = - ------
    C(n,k)   k C(n,k)

Suppose [10] is true for some m, then

    ---     j C(m+1,j)
    >   (-1)  --------
    ---       C(n+j,k)
     j

      ---     j    C(m,j)    C(m,j-1)
    = >   (-1)  ( -------- + -------- )
      ---         C(n+j,k)   C(n+j,k)
       j

      ---     j  C(m,j)    ---     j   C(m,j)
    = >   (-1)  -------- - >   (-1)  ----------
      ---       C(n+j,k)   ---       C(n+j+1,k)
       j                    j

       k       1        k        1
    = --- ---------- - --- ------------
      k+m C(n+m,k+m)   k+m C(n+m+1,k+m)

       k   k+m         1
    = --- ----- --------------
      k+m k+m+1 C(n+m+1,k+m+1)

        k          1
    = ----- --------------
      k+m+1 C(n+m+1,k+m+1)

Thus, it is true for m+1.  Therefore, [10] is true for all m >= 0.