theorem
  i is odd implies i |^ k is odd
proof
  defpred P[Nat] means i |^ $1 is odd;
  assume
A1: i is odd;
A2: for n holds P[n] implies P[n+1]
  proof
    let n;
A3: i |^ (n+1) = i |^ n * i by NEWTON:6;
    assume i |^ n is odd;
    hence thesis by A1,A3;
  end;
  i |^ 0 = 1 & 1 mod 2 = 1 by NAT_D:24,NEWTON:4;
  then
A4: P[0] by NAT_2:22;
  for n holds P[n] from NAT_1:sch 2(A4,A2);
  hence thesis;
end;
