I don't seem to grasp the proof. First we construct a vector space over a subfield with prime order $p$ where $p$ is the characteristic of the field . As the field is finite , the vector space will be finite-dimensional so there is a finite basis of $n$ elements that spans the space and every vector in that space can be uniquely expressed as a linear combination of the basis , so the field has order $p^n$. My question is : why can't we just take any subfield , for example the field itself , as the underlying field and conclude that our field $F$ has $|F|^n$ elements?
[Math] The proof that a finite field has a prime power order
abstract-algebrafinite-fieldsring-theory
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Let me address your first question. First, I want to argue that there is no precise meaning of "involving numbers only". For example, given a finite field $F$ of size $4$ constructed in the usual manner (quotient of a polynomial ring over $\mathbb{Z}/2\mathbb{Z}$), I can choose a set of numbers, say $$S=\{37,\tfrac{5}{19},\pi,e\}$$ and, choosing a bijection of $S$ with $F$, use transport of structure to give $S$ the structure of a field. The field structure does not depend in any way on what the underlying set is "made of".
However, along the lines of what I think you are ultimately after, you can obtain finite fields of any possible order using larger rings of integers. For example, $\mathbb{Z}[i]/(3)$ is a finite field of size $9$, and $\mathbb{Z}[i]$ consists of very reasonable numbers, $$\mathbb{Z}[i]=\{a+bi\mid a,b\in\mathbb{Z}\}.$$
Now let me addressr your second question. Let's use $\mathbb{F}_p$ to mean $\mathbb{Z}/p\mathbb{Z}$, a finite field of order $p$ - it is a very common notation that is slightly less cumbersome, but doesn't mean anything different, they are exact synonyms.
A finite field of order $p^n$ is often constructed by taking the polynomial ring $\mathbb{F}_p[x]$, choosing an irreducible polynomial $f\in \mathbb{F}_p[x]$ of degree $n$, and then making the field $$F=\mathbb{F}_p[x]/(f).$$ Now, the division algorithm for polynomials tells you that each equivalence class in this quotient can be uniquely identified by a representative of degree $<n$. In other words, $$\begin{align*} F&=\{a_0+a_1x+\cdots +a_{n-1}x^{n-1}+(f)\mid a_0,a_1,\ldots,a_{n-1}\in\mathbb{F}_p\}\\\\ &=\left\{\,\overline{a_0+a_1x+\cdots +a_{n-1}x^{n-1}}\,\;\middle\vert\;a_0,a_1,\ldots,a_{n-1}\in\mathbb{F}_p\right\}\\\\\\ &=\{a_0+a_1\overline{x}+\cdots +a_{n-1}\overline{x}^{n-1}\mid a_0,a_1,\ldots,a_{n-1}\in\mathbb{F}_p\} \end{align*}$$ Letting the symbol $\alpha$ be a stand-in for $\overline{x}$, you can think of $F$ as being $\mathbb{F}_p$ with a new element "$\alpha$" added in, where $\alpha$ is a root of $f$, and you can write $F=\mathbb{F}_p[\alpha]$.
Now, the prime subfield of $F$ is just the "constant" polynomials, i.e. the ones with no $\alpha$'s in them: $$\text{the prime subfield of }F=\{a_0+0\alpha+\cdots+0\alpha^{n-1}\mid a_0\in\mathbb{F}_p\}$$ and for each divisor $d\mid n$, the unique subfield of $F$ of order $p^d$ is the collection of polynomials in $\alpha$ whose terms are those of exponents that are multiples of $n/d$: $$\text{the subfield of }F\text{ of order }p^d=\{a_0+a_1\alpha^{n/d}+\cdots+a_{d-1}\alpha^{(d-1)n/d}\mid a_0,a_1,\ldots,a_{d-1}\in\mathbb{F}_p\}$$ (clearly, the above set has cardinality $p^d$, because it takes $d$ elements of $\mathbb{F}_p$ to specify a given element of the above set, namely, each of the coefficients of the powers of $\alpha$. To see that it is a field, remember that $(a+b)^p=a^p+b^p$ in a field of characteristic $p$.)
An important result of linear algebra tells us that two $K$-vector spaces are isomorphic if and only if they have the same dimension.
As $m = \dim_K(L)$ by definition and $m = \dim_K(V_m(K))$ (for example by considering the standard basis), you get $L \cong V_m(K)$ as $K$-vector spaces.
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The dimension of a vector space depends on the field you're talking about, so thinking its dimension over the prime subfield and its dimension over itself are the same is a mistake! (Or a typo: sorry if that's the case :) )
Choosing $F$ itself would not be very helpful, since $F$ is one dimensional over $F$, and thus $n=1$, so that your result is $|F|=|F|^1$.
You can use other subfields: if $K$ is a subfield of $F$ and $F$ is $n$ dimensional over $K$, then $|F|=|K|^n$. And then if $E$ is a subfield of $K$ over which $K$ is $m$ dimensional, $|K|=|E|^m$, and $|F|=|E|^{mn}$. It's all consistent.
The nice part about using the prime subfield is that we're guaranteed every finite field contains a copy of one of the prime order fields. It's just a nice, canonical choice.