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Proof. Note that $P'$ has degree $< \deg (P)$. Hence if $P$ and $P'$ are not relatively prime, then $(P, P') = (R)$ where $R$ is a polynomial of degree $< \deg (P)$ contradicting the irreducibility of $P$. This proves we have the dichotomy between (1) and (2).

Assume we are in case (2) and $P = a_ d x^ d + \ldots + a_0$. Then $P' = da_ d x^{d - 1} + \ldots + a_1$. In characteristic $0$ we see that this forces $a_ d, \ldots , a_1 = 0$ which would mean $P$ is constant a contradiction. Thus we conclude that the characteristic $p$ is positive. In this case the condition $P' = 0$ forces $a_ i = 0$ whenever $p$ does not divide $i$. In other words, $P(x) = P_1(x^ p)$ for some nonconstant polynomial $P_1$. Clearly, $P_1$ is irreducible as well. By induction on the degree we see that $P_1(x) = Q(x^ q)$ as in the statement of the lemma, hence $P(x) = Q(x^{pq})$ and the lemma is proved. $\square$