
theorem Th3:
  for a,n being Nat st a > 1 holds a|^n > n
proof
  defpred P[Nat] means for a being Nat st a>1 holds a|^$1 > $1;
A1: for n being Nat holds P[n] implies P[n+1]
  proof
    let n be Nat;
    assume
A2: P[n];
    now
      let a be Nat;
      assume
A3:   a>1;
      then a>=1+1 by NAT_1:13;
      then
A4:   a*n>=2*n by XREAL_1:64;
      per cases;
      suppose
        n=0;
        hence a|^(n+1)>n+1 by A3;
      end;
      suppose
        n>0;
        then n>=0+1 by NAT_1:13;
        then n+n>=n+1 by XREAL_1:6;
        then
A5:     a*n>=n+1 by A4,XXREAL_0:2;
        (a|^n)*a>n*a by A2,A3,XREAL_1:68;
        then a|^(n+1)>n*a by NEWTON:6;
        hence a|^(n+1)>n+1 by A5,XXREAL_0:2;
      end;
    end;
    hence thesis;
  end;
A6: P[0];
  for n being Nat holds P[n] from NAT_1:sch 2(A6,A1);
  hence thesis;
end;
