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

theorem Th38:
  p is prime & d > 1 & d divides (p |^ k) & not d divides ((p |^ k) div p)
  implies d = p |^ k
proof
  assume that
A1: p is prime and
A2: d > 1 and
A3: d divides (p |^ k) and
A4: not d divides ((p |^ k) div p);
A5: k <> 0
  proof
    assume k = 0;
    then p |^ k = 1 by NEWTON:4;
    hence contradiction by A2,A3,NAT_D:7;
  end;
  then k >= 1 by NAT_1:14;
  then
A6: k - 1 >= 1 - 1 by XREAL_1:9;
  consider t being Element of NAT such that
A7: d = p |^ t and
A8: t <= k by A1,A3,Th34;
A9: p <> 0 by A1,INT_2:def 4;
  not t < k
  proof
    assume t < k;
    then t < k + -1 + 1;
    then t < k -'1 + 1 by A6,XREAL_0:def 2;
    then t <= k -'1 by NAT_1:13;
    then d divides p |^ (k -'1) by A7,Lm6;
    hence contradiction by A4,A9,A5,Th28,NAT_1:14;
  end;
  hence thesis by A7,A8,XXREAL_0:1;
end;
