theorem Lm4:
  for n be Nat holds (1.A)|^n = 1.A
  proof
    defpred P[Nat] means (1.A)|^$1 = 1.A;
A1: now
      let n be Nat;
      assume
A2:   P[n];
      (1.A)|^(n+1) = (1.A)|^n * (1.A)|^1 by BINOM:10
        .= 1.A by BINOM:8,A2;
      hence P[n+1];
    end;
    (1.A)|^0 = 1_A by BINOM:8; then
A3: P[0];
    thus for n be Nat holds P[n] from NAT_1:sch 2(A3,A1);
  end;
