theorem Th36:
  g.(a |^ n) = (g.a) |^ n
proof
  defpred Q[Nat] means g.(a |^ $1) = (g.a) |^ $1;
A1: for n being Nat st Q[n] holds Q[n + 1]
  proof
    let n be Nat;
    assume
A2: Q[n];
    thus g.(a |^ (n + 1)) = g.(a |^ n * a) by GROUP_1:34
      .= (g.a) |^ n * g.a by A2,Def6
      .= (g.a) |^ (n + 1) by GROUP_1:34;
  end;
  g.(a |^ 0) = g.(1_G) by GROUP_1:25
    .= 1_H by Th31
    .= (g.a) |^ 0 by GROUP_1:25;
  then
A3: Q[0];
  for n being Nat holds Q[n] from NAT_1:sch 2(A3,A1);
  hence thesis;
end;
