You should observe that $z^{p^r}-z=0$ for all $z\in\Bbb{F}_q$. Also this is the lowest degree polynomial vanishing at all the elements, because there are $p^r$ elements in all. You can reinterpret this as stating that $x^r-1$ is the minimal polynomial of $L$. As the characteristic polynomial of $L$ is of degree $r$, it must coincide with the minimal polynomial. Thus as a module over $\Bbb{F}_p[x]$ the extension field is cyclic, and we can conclude that
$$
\Bbb{F}_{p^r}\cong \Bbb{F}_p[x]/\langle x^r-1\rangle.
$$
Beware that this isomorphism is only as modules. The latter object also has natural structure as a ring, but this is not an isomorphism of rings.
Another point of view is the following.
The extension $\Bbb{F}_q/\Bbb{F}_p$ has a so called normal basis. That is a basis $\{x_1,x_2,\ldots,x_r\}$ of the form $x_i=L^{i-1}(x_1)=x_1^{p^{i-1}}$ for all $i=2,3,\ldots,r$. With respect to such a basis the matrix of $L$ is a simple permutation matrix of an $r$-cycle. This leads to the same conclusion and allows us to describe the above isomorphism $\Phi: \Bbb{F}_p[x]/\langle x^r-1\rangle\to \Bbb{F}_{p^r}$ as
$$
\Phi(f(x))=f(L)\cdot x_1.
$$