I've come up with a very odd proof that the polynomial $c_0 + c_1x^p + c_2x^{2p} + \dots + c_nx^{np}$ is always reducible in a perfect field of characteristic $p$. Honestly, the more I look at this proof, the more I don't believe in it, so I'm hoping for a more direct approach here.
Let $F$ be a field of characteristic $p$ and let $f(x) \in F[x]$ be defined by $f(x) = c_0 + c_1x^p + c_2x^{2p} + \dots + c_nx^{np}$ for some $n \in \mathbb{N}$. Suppose that $f$ is irreducible. As $f'(x) = 0$, $f$ is not separable in $F$.
Define $g(x) \in F[x]$ by $g(x) = c_0 + c_1x + c_2x^2 + \dots + c_nx^n$. As $f(x) = g(x^p)$, if $g$ was reducible over $F$, $f$ would be as well. Therefore, $g$ is irreducible. Note that $g'(x) = c_1 + 2c_2x + \dots + nc_nx^{n - 1}$. This is only $0$ if every coefficient of $g$ next to a power of $x$ that is not a multiple of $x^p$ is $0$. If the derivative is $0$, define $h(x) = c_0 + c_1x^{\frac{1}{p}} + \dots + c_nx^{\frac{1}{p}}$. Here $h$ is a polynomial by assumption and is irreducible by the same argument as for $g$. Therefore, we continue inductively until, in a finite number of steps, we arrive at a polynomial $k(x)$ which is also irreducible and the derivative of which is nonzero.
Let $E$ be a splitting field for $k$. As $k'(x) \neq 0$, $k$ is separable and so has distinct roots in $E$. As $E$ is an algebraic extension of $F$, $E$ is also perfect and so there exists an automorphism $\varphi(x) = x^p$ on $E$. In particular, this means that every root of $k$ corresponds to exactly one root of $f$. As such, $f$ also has distinct roots on $E$ and is therefore separable, contradicting the fact that $f'(x) = 0$.
This looks both like black magic and so incredibly overcomplicated that there has to be an error in there somewhere. I've checked it five times by now, though, and I can't find anything. Please let me know if you find an error or if you know of a simpler approach.