[proofplan]
We first rule out characteristic $0$ by observing that the integer multiples of the identity would give infinitely many distinct elements of $k$. The characteristic dichotomy for fields then gives $\operatorname{char}(k)=p$ for a prime number $p$, and the prime subfield classification identifies the prime subfield of $k$ with $\mathbb{F}_p$. Finally, we regard $k$ as a finite-dimensional [vector space](/page/Vector%20Space) over this copy of $\mathbb{F}_p$ and count the coordinate tuples relative to a basis.
[/proofplan]
[step:Exclude characteristic $0$ using finiteness]
Let
\begin{align*}
\iota: \mathbb{Z} &\to k
\end{align*}
\begin{align*}
m &\mapsto m \cdot 1_k
\end{align*}
be the canonical ring homomorphism from $\mathbb{Z}$ to $k$, where $m \cdot 1_k$ denotes repeated addition of $1_k$ if $m>0$, $0_k$ if $m=0$, and repeated addition of $-1_k$ if $m<0$.
Suppose for contradiction that $\operatorname{char}(k)=0$. By the definition of characteristic, $\ker(\iota)=\{0\}$, so $\iota$ is injective. Therefore the subset $\iota(\mathbb{Z}) \subset k$ is infinite. This contradicts the hypothesis that $k$ is finite. Hence $\operatorname{char}(k) \neq 0$.
[guided]
The obstruction to characteristic $0$ is that characteristic $0$ forces an infinite copy of the integers inside the field. Define the canonical ring homomorphism
\begin{align*}
\iota: \mathbb{Z} &\to k
\end{align*}
\begin{align*}
m &\mapsto m \cdot 1_k.
\end{align*}
Here $m \cdot 1_k$ means adding the multiplicative identity $1_k$ to itself $m$ times when $m>0$, taking $0_k$ when $m=0$, and adding $-1_k$ to itself $|m|$ times when $m<0$.
If $\operatorname{char}(k)=0$, then no positive integer multiple of $1_k$ is equal to $0_k$. Equivalently, the kernel of $\iota$ is exactly $\{0\}$. Since a ring homomorphism with zero kernel is injective, $\iota$ is injective. Thus $\iota(\mathbb{Z})$ has the same cardinality as $\mathbb{Z}$ and is an infinite subset of $k$. This is impossible because $k$ is finite. Therefore $\operatorname{char}(k)$ cannot be $0$.
[/guided]
[/step]
[step:Use the field characteristic dichotomy to obtain a prime characteristic]
By [citetheorem:8303], the characteristic of a field is either $0$ or a prime number. The previous step shows that $\operatorname{char}(k)\neq 0$. Hence there exists a prime number $p$ such that
\begin{align*}
\operatorname{char}(k)=p.
\end{align*}
[/step]
[step:Identify the prime subfield with $\mathbb{F}_p$]
Let $k_0 \subset k$ denote the prime subfield of $k$, namely the smallest subfield of $k$ containing $1_k$. Since $\operatorname{char}(k)=p$, [citetheorem:8304] gives a field isomorphism
\begin{align*}
\varphi: \mathbb{F}_p &\to k_0.
\end{align*}
We use this isomorphism to define scalar multiplication of $\mathbb{F}_p$ on $k$ by
\begin{align*}
\mathbb{F}_p \times k &\to k
\end{align*}
\begin{align*}
(\lambda, x) &\mapsto \varphi(\lambda)x,
\end{align*}
where the product on the right is multiplication in the field $k$. Since $\varphi$ is a field homomorphism and the addition and multiplication laws are inherited from $k$, the field axioms of $k$ imply the vector space axioms. Thus $k$ is a vector space over $\mathbb{F}_p$.
[/step]
[step:Choose a finite basis over the prime field]
Because $k$ is a finite set, every linearly independent subset of $k$ over $\mathbb{F}_p$ is finite. Choose a linearly independent subset $B \subset k$ over $\mathbb{F}_p$ of maximal cardinality, which exists because the collection of linearly independent subsets of the finite set $k$ is finite and nonempty. Let
\begin{align*}
n := |B|.
\end{align*}
The set $B$ spans $k$ over $\mathbb{F}_p$: if some $x \in k$ were not in the $\mathbb{F}_p$-span of $B$, then $B \cup \{x\}$ would still be linearly independent, contradicting the maximality of $B$. Therefore $B$ is a basis of the $\mathbb{F}_p$-vector space $k$, and
\begin{align*}
n=\dim_{\mathbb{F}_p}(k).
\end{align*}
Since $1_k \neq 0_k$, the vector space $k$ is not the zero vector space, so $n \geq 1$ and hence $n \in \mathbb{N}$.
[/step]
[step:Count coordinate tuples relative to the basis]
Write the basis as
\begin{align*}
B=\{b_1,\dots,b_n\}.
\end{align*}
Define the coordinate map
\begin{align*}
\Psi: \mathbb{F}_p^n &\to k
\end{align*}
\begin{align*}
(\lambda_1,\dots,\lambda_n) &\mapsto \varphi(\lambda_1)b_1+\dots+\varphi(\lambda_n)b_n.
\end{align*}
Because $B$ spans $k$ over $\mathbb{F}_p$, the map $\Psi$ is surjective. Because $B$ is linearly independent over $\mathbb{F}_p$, the map $\Psi$ is injective. Hence $\Psi$ is a bijection, so
\begin{align*}
|k|=|\mathbb{F}_p^n|.
\end{align*}
Since $\mathbb{F}_p$ has exactly $p$ elements, the Cartesian product $\mathbb{F}_p^n$ has cardinality
\begin{align*}
|\mathbb{F}_p^n|=p^n.
\end{align*}
Therefore
\begin{align*}
|k|=p^n.
\end{align*}
This proves the stated prime-power form of the cardinality, with $p=\operatorname{char}(k)$ and $n=\dim_{\mathbb{F}_p}(k)$.
[/step]
