= Solution
The <degree of a field extension> $[K:M]$ is the <dimension of a vector space> of $K$ over $M$. If $(\alpha_i)$ is an $M$-basis of $K$ and $(\beta_j)$ a $K$-basis of $L$, then $(\alpha_i\beta_j)$ spans $L$ over $M$. An $M$-linear relation among these products, regrouped by $j$, first vanishes coefficientwise by independence of the $\beta_j$, and then by independence of the $\alpha_i$. This proves the <tower law> $[L:M]=[L:K][K:M]$, including cardinal dimensions.
A <finite field> has prime <characteristic> $p$ and is a finite-dimensional <vector space> over its prime field $\mathbb F_p$. Thus its size is $p^n$. A subfield of size $p^m$ forces $m\mid n$ by the <tower law>. Conversely, if $m\mid n$, the polynomial $X^{p^m}-X$ divides $X^{p^n}-X$; all its $p^m$ distinct roots therefore lie in $K$. They are closed under addition, multiplication and inverses by the <Finite-field Frobenius automorphism>, forming the required subfield.
For an <irreducible polynomial> of degree $d$ over $\mathbb F_q$, a root generates $\mathbb F_{q^d}$. Its conjugates are $\alpha,\alpha^q,\ldots,\alpha^{q^{d-1}}$, so this field is already the <splitting field>. The automorphism $x\mapsto x^q$ cycles these roots and has order $d$. Consequently
$$
\boxed{\operatorname{Gal}(f/\mathbb F_q)\cong C_d.}
$$
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