= Solution
If the <minimal polynomial> of $\gamma$ has <degree of a polynomial>[degree] $d$, then
$\mathbb F_p(\gamma)$ has $p^d$ elements and is an <intermediate field> of
$L/\mathbb F_p$. The <tower law for field extensions> gives
$$
n=[L:\mathbb F_p]
=[L:\mathbb F_p(\gamma)]d,
$$
so $d\mid n$.
The <multiplicative group of a finite field is cyclic>. Choose a generator
$\gamma$ of $L^\times$, of <order of a group element>[order] $p^n-1$. If $\gamma$ lay in a proper <intermediate field> of degree $d<n$, its order would <divisibility>[divide] $p^d-1<p^n-1$, a contradiction. Thus
$\mathbb F_p(\gamma)=L$ and its <minimal polynomial> has degree $n$.
For arbitrary $r\geq1$, let $E$ be the <splitting field> of
$X^{p^r}-X$ over $\mathbb F_p$. Its <roots of a polynomial>[roots] form a <field>: the <Frobenius endomorphism> shows they are closed under <addition>, <subtraction>, <multiplication>, and <multiplicative inverse>[inversion]. The <derivative> is $-1$, so there are exactly $p^r$ distinct roots. Thus this root field has order $p^r$. Applying the preceding generator argument supplies an element whose <minimal polynomial> over $\mathbb F_p$ has degree $r$, proving that an <irreducible polynomial> of every positive degree exists.
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