
theorem thX0:
for p being Prime
for n being non zero Nat
for F being Field st card F = p|^n
for a being Element of F holds a|^(p|^n) = a
proof
let p be Prime, n be non zero Nat, F be Field;
assume A: card F = p|^n;
let a be Element of F;
reconsider F as finite Field by A;
set M = MultGroup F;
reconsider n1 = n - 1 as Element of NAT by INT_1:3;
H: the carrier of F = (the carrier of M) \/ {0.F} by UNIROOTS:15;
per cases;
suppose a <> 0.F;
  then not a in {0.F} by TARSKI:def 1;
  then reconsider b = a as Element of M by H,XBOOLE_0:def 3;
  card M = p|^n - 1 by A,UNIROOTS:18; then
  b * (b|^(p|^n - 1)) = b * 1_M by GR_CY_1:9 .= a by GROUP_1:def 4;
  hence a = b|^(1 + (p|^n - 1)) by GROUP_1:34
         .= a|^(p|^n) by FIELD_15:10;
  end;
suppose a = 0.F;
  hence a|^(p|^n) = a;
  end;
end;
