Stabilizer of a Group Action

Here's another approach, based on the

Lemma: Let $G$ be a finite $p$-group, acting on the finite set $S$. If $F$ is the set of fixed points of this action, then $$ |F|\equiv|S|\pmod{p}$$

The proof is easy, and is really just the fact that an orbit of size bigger than $1$ has to have size dividing $G$, and hence is $0\pmod{p}$.

Now let $G=\langle L\rangle$, and $S=\mathbb{F}_p^n$, so that $F=\{v\in\mathbb{F}_p^n\mid L(v)=v\}$. By the above lemma, $$ |F|\equiv|S|\equiv0\pmod{p}$$

Since $0\in F$, we see that $|F|>0$, so that $|F|\ge p$. Thus, there is a nonzero vector $v\in F$.

Hint: In the ring $\mathrm{Hom}_{\mathbb F_p}(\mathbb F_p^n,\mathbb F_p^n)$, we have $L^{p^k}-1=(L-1)^{p^k}$.