$f(X^p)$ irreducible or $p$th power if $f$ irreducible An exercise in Bourbaki:

Let $K$ be a field of characteristic $p>0$ and $f$ irreducible monic polynomial of $K[X]$. Show that in $K[X]$ the polynomial $f(X^p)$ is either irreducible or the $p$th power of an irreducible polynomial, depending of whether or not there exists a coefficient of $f$ not belonging to $K^p$.

The given suggestion is to decompose $f(X^p)$ in linear factors in an algebraic closure of K.
Thus far I have done this: I take $\theta$ to be a root of $f(X^p)$. By considering the subextions $K(\theta ^p)/K$ of $K(\theta)/K$, we see that $f$ is irreducible iff  the degree of $K(\theta ^p)/K(\theta)$ is $p$ to $K(\theta^p)$, iff $K(\theta)=K(\theta^p)$.
 A: I did not manage to give a proof considering those field extensions, so I hope you don't mind if my suggestion for a proof uses another method:
Let $f(x) = (x-\theta_1)\dots(x-\theta_n)$ in an algebraic closure $\overline K$ of $K$. 
Let $\lambda_i$ be the unique element of $\overline K$ with $\lambda_i ^p=\theta_i$.
Then $$f(x^p) = (x^p-\theta_1) \dots (x^p-\theta_n) =(x^p-\lambda_1^p) \dots (x^p-\lambda_n^p) = (x-\lambda_1)^p \dots (x-\theta_n)^p=\Big ((x-\lambda_1) \dots (x-\lambda_n) \Big )^p.$$
Now let $(x-\lambda_1) \dots (x-\lambda_n) = x^n+a_nx^{n-1}+...+a_0 \in \overline K[X]$, then we find 
$$f(x^p) = (x^n)^{p}+a_n^p(x^{n-1})^p+...+a_0^p.$$
Now suppose all coefficients of $f(x^p)$ are already in $K^p$, then all $a_i$ are in $K$ and thus $g = (x-\lambda_1) \dots (x-\lambda_n)\in K[X].$ 
For showing that $g$ is irreducible it is sufficient to see that if $(x-\lambda_1) \dots (x-\lambda_m)$ with $m<n$ was already in $K[X]$, then so would be $(x-\lambda_1^p) \dots (x-\lambda_m^p)=(x-\theta_1) \dots (x-\theta_m)$ which is a contradiction to irreducibility of $f$. This completes one direction of our proof.
Suppose otherwise there is a coefficient of $g$ not belonging to $K^p$. We show, that $f(x^p)$ is irreducible:
Suppose that is not the case. We know, that $f(x^p)$ is not of the form $g^p$. However, the factors of $f(x^p) = g^p \cdot h^p$ with some non-trivial $g^p,h^p \in K[X]$ have to be $p$-th powers too, because they must not share any roots in $\overline K$. So $g,h$ are of the form $g'(x^p),h'(x^p)$ with $g',h' \in K[X]$, showing $g'(x)\cdot h'(x)=f(x)$, a contradiction to irreducibility of $f$.
