reserve d,i,j,k,m,n,p,q,x,k1,k2 for Nat,
  a,c,i1,i2,i3,i5 for Integer;

theorem Th26:
  n <> 0 & p <> 0 implies n |^ p = n*(n |^ (p -'1))
proof
  assume
A1: n <> 0 & p <> 0;
  defpred P[Nat] means n <> 0 & $1 <> 0 implies n |^ $1 = n*(n |^ ($1 -' 1));
A2: for m holds P[m] implies P[m+1]
  proof
    let m;
    assume P[m];
    now
      per cases;
      suppose
A3:     m = 0;
        n*(n |^ ((0+1) -' 1)) = n*(n |^ 0) by XREAL_1:232
          .= n*1 by NEWTON:4;
        hence thesis by A3;
      end;
      suppose
A4:     m <> 0;
        n |^ (m+1) = (n |^ m) * n by NEWTON:6
          .= (n |^ (m -' 1 + 1))*n by A4,NAT_1:14,XREAL_1:235
          .= n*(n |^ ((m + 1) -' 1)) by A4,Lm4;
        hence thesis;
      end;
    end;
    hence thesis;
  end;
A5: P[0];
  for m holds P[m] from NAT_1:sch 2(A5,A2);
  hence thesis by A1;
end;
