Any finite field $K$ with characteristic $p$ is isomorphic to $\mathbb{Z}_p[t]/(f)$ for some irreducible $f\in\mathbb{Z}_p[x]$. (From now on, assume $K=\mathbb{Z}_p[t]/(f)$.) Since $p\mid\dbinom{p}{k}$ for $1\le k\le p-1$ we have that $(a+b)^p=a^p+b^p$ for all $a, b\in K$, so if $f=\displaystyle\sum_{i=0}^na_ix^i$ then by FLT we have $$f(t)^p=\left(\displaystyle\sum_{i=0}^na_ix^i\right)^p=\displaystyle\sum_{i=0}^na_i^px^p=\displaystyle\sum_{i=0}^na_ix^p=f(x^p).$$ Since $t\in K$ is a root of $f$, so are $t^p$, $t^{p^2}$, ... $t^{p^{n-1}}$.
By Vieta, the sum of the roots of $f$ equals $-\dfrac{a_1}{a_0}$, so $$t+t^p+t^{p^2}+...+t^{p^{n-1}}=-\dfrac{a_1}{a_0}.$$ So $t$ is a root of $$g=x^{p^{n-1}}+x^{p^{n-2}}+...+x^p+x+\dfrac{a_1}{a_0}.$$ Since $f$ is irreducible, $f$ is the minimal polynomial of $t$, so $f\mid g$ in $\mathbb{Z}_p[x]$. If we fix $r=\dfrac{a_1}{a_0}$ and let $a_0$, $a_1$, ..., $a_n$ range so that $f$ is irreducible, then $g$ stays fixed, so we can find some irreducible factors of $g$.
For example, if $f=x^2-ax+b$, then $f\mid x^p+x-a$ in $\mathbb{Z}_p[x]$. If $p=2$ you get $x^2+x+1 \mid x^2+x+1$, duh. So assume $p$ is odd. Then for a fixed $a$, there are $\dfrac{p+1}{2}$ different $b$'s that will make $x^2-ax+b$ reducible: if $r$ is a root, then $b=r(a-r)$, and two different $r$'s give the same $b$ if and only if they sum to $a$. So $r=\dfrac{a}{2}$ has no partner while the remaining $p-1$ guys each have a partner, giving $1+\dfrac{p-1}{2}=\dfrac{p+1}{2}$ distinct $b$'s. So there are $\dfrac{p-1}{2}$ $b$'s that will make $x^2-ax+b$ irreducible, and each of these $x^2-ax+b$ guys divides $x^p+x-a$. That's already a degree of $p-1$, so there's just one irreducible factor left: a linear. And for any $r\in\mathbb{Z}_p$, $r^p+r=a\iff 2r=a\iff r=\dfrac{a}{2}$ by FLT, so the last factor is $\left(x-\dfrac{a}{2}\right)$. So we've just factored $x^p+x-a$ into irreducibles over $\mathbb{Z}_p[x]$.
We can of course do this with the other symmetric sums.
Is my logic correct? Where can I read more about this technique?