you do know that in a field F with char(F) = p, that:
in particular, is the ONLY root of , in any extension of .
so it suffices to show .
for suppose it were. then we would have:
for some polynomials .
suppose deg(p) = k, and deg(q) = m. then deg(p(x)p) = kp, and deg(q(x)p) = mp.
thus we have:
1 + mp = kp
but p divides kp, and cannot divide 1 + mp. so we have our contradiction.