you do know that in a field F with char(F) = p, that:

, right?

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.