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

theorem Th25:
  m > 1 & n > 0 implies m |^ n > 1
proof
  assume that
A1: m > 1 and
A2: n > 0;
  defpred P[Nat] means $1 > 0 implies m |^ $1 > 1;
A3: for n holds P[n] implies P[n+1]
  proof
    let n;
A4: m |^ (n+1) = (m |^ n)*(m |^ 1) by NEWTON:8
      .= (m |^ n)*m;
    assume
A5: P[n];
    P[n+1]
    proof
      now
        per cases;
        suppose
          n = 0;
          hence thesis by A1;
        end;
        suppose
          n <> 0;
          then (m |^ n)*m > 1*m by A1,A5,XREAL_1:68;
          hence thesis by A1,A4,XXREAL_0:2;
        end;
      end;
      hence thesis;
    end;
    hence thesis;
  end;
A6: P[0];
  for n holds P[n] from NAT_1:sch 2(A6,A3);
  hence thesis by A2;
end;
