= Solution
Put $F=L^G$ and $n=|G|$. The <automorphism-count bound for a finite field extension> gives
$$
n\leq[L:F].
$$
For the reverse inequality, let $x_1,\ldots,x_m$ be linearly independent over $F$. We claim that the $n\times m$ matrix
$$
M=(g(x_j))_{g\in G,\,1\leq j\leq m}
$$
has linearly independent columns over $L$. If not, choose a nonzero relation
$$
\sum_jc_jg(x_j)=0\qquad\text{for every }g\in G
$$
with the fewest nonzero coefficients, and normalize one coefficient to $c_1=1$. Applying $h\in G$ and reindexing the rows gives another relation with coefficients $h(c_j)$. Subtracting eliminates the first term, so minimality forces $h(c_j)=c_j$ for every $h$ and $j$. Thus every $c_j\in F$. The row for the identity automorphism then contradicts the $F$-linear independence of the $x_j$.
Therefore $m\leq n$. Taking an $F$-basis of $L$ gives $[L:F]\leq n$, and hence
$$
\boxed{[L:L^G]=|G|.}
$$
This proves the degree assertion in the <Artin fixed-field theorem>.
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