Go up to a splitting field $E$ so that in there $f(x)$ splits completely. Now a root of $f(x)$ is $\sqrt[p^n]{c}$ and so we can write $f(x)$ over $E$ as
$f(x) = x^{p^n} - (\sqrt[p^n]{c})^{p^n} = (x - \sqrt[p^n]{c})^{p^n}$
by the schoolboy binomial theorem. Now recall $f(x) \in \Bbb{F}[x]$ so that each coefficient of $f(X)$ is an element of $\Bbb{F}$. Suppose that $f(x)$ is reducible. From the factorisation above, we see that any divisor of $f(x)$ must be of the form
$(x - \sqrt[p^n]{c})^{r}$
where $0 \leq r \leq p^n$. Now expand this out by the ordinary binomial theorem and look at the coefficient of the second highest power of the indeterminate $x$. Can you deduce the irreducibility of $f(x)$ from here? Hint: show that $r = 0$ or $r = p^n$.