reserve a,b,i,j,k,l,m,n for Nat;

theorem
  for a be non zero Integer st 1 <= k <= n holds
    a divides ((a,b) Subnomial n).k
proof
  let a be non zero Integer;
  A0: len ((a,b) Subnomial n) = n+1 by Def2;
  assume
  A1: 1<= k <= n; then
  reconsider m = k-1 as Nat;
  (k-1)+1 > (k-1)+0 by XREAL_1:6; then
  consider l be Nat such that
  A2: n = m+l by A1,XXREAL_0:2,NAT_1:10;
  m+l >= m+1 by A1,A2; then
  l >= 1 by XREAL_1:6; then
  reconsider s = l-1 as Nat;
  A4: l = n - m by A2;
  k+0 <= n+1 by XREAL_1:7,A1; then
  k in dom ((a,b) Subnomial n) by A0,A1,FINSEQ_3:25; then
  ((a,b) Subnomial n).k = a|^(s+1)*b|^m by Def2,A4
  .= a*a|^s*b|^m by NEWTON:6
  .= a*(a|^s*b|^m);
  hence thesis;
end;
