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

theorem
  for a be Real, n be Nat holds
    a|^(n+1) = Sum((a,1) Subnomial n)*(a-1) + 1
proof
  let a be Real, n be Nat;
  a|^(n+1) - 1|^(n+1) = (a-1)*Sum ((a,1) Subnomial n) by SumS;
  hence thesis;
end;
