I’ll explain how to find the number of monic irreducibles of degree $6$ over $\Bbb F_5$, and I hope that the general method will be obvious from this example. On the other hand, finding one of them is another kettle of fish.
For ease in typing, I’ll call $k_n$ the field with $5^n$ elements. Then $\Bbb F_5=k_1$, $\Bbb F_{25}=k_2$, etc. I want to consider $k_6$. It has three proper subfields, namely $k_1$, $k_2$, and $k_3$. Let’s let $N_m$ be the number of elements $\alpha$ of $k_6$ with the property that $\alpha\in k_m$ but in no proper subfield of $k_m$.
Let’s notice two things: first, that we have gotten a partition of the set $k_6$ into four disjoint sets, so that $N_1+N_2+N_3+N_6=5^6$, and second, that such an element $\alpha\in k_m$ not in a proper subfield has exactly $m$ conjugates (including itself) over $\Bbb F_5$, and the monic polynomial whose roots are those conjugates will be an $\Bbb F_5$-irreducible polynomial of degree $m$. Furthermore, if we call $I_m$ the number of monic irreducibles of degree $m$, we have $mI_m=N_m$. Therefore, we have:
$$
\begin{align}
5^6&=N_1+N_2+N_3+N_6&\text{and more generally}\\
5^m&=\sum_{d|m}N_d&\text{for any $m$}\\
N_m&=\sum_{d|m}\mu(d)5^{m/d}=mI_m\,,
\end{align}
$$
where $\mu$ is the Möbius function. For our special case $\mu(1)=\mu(6)=1$ and $\mu(2)=\mu(3)=-1$. In particular, the number of irreducible monic sextics over $\Bbb F_5$ is
$$
I_6=\frac16\bigl(5^6-5^3-5^2+5\bigr)=2580
$$
