Why $F^S$ represents set of functions from S to F? This is a definition from the book "Linear Algebra Done Right"
$F^S$ Notation


*

*If $S$ is a set, then $F^S$ denotes the set of functions from $S$ to $F$.

*And $F^S$ is a vector space with following addition and scalar multiplication.

*$\forall f, g \in F^S, \ \exists f + g \in F^S$ is a function defined by
$$\forall x \in S, \ (f + g)(x) = f(x) + g(x)$$

*$\forall \lambda \in F, \ \forall f \in F^S, \ \exists \lambda f \in F^S$ is the function defined by
$$\forall x \in S, \ (\lambda f )(x) = \lambda f(x)$$

Here $F$ denotes a field.
My Question is why $F^S$, I don't understand the intuition behind this notation. Everything work perfectly fine even if we say

Let $X$ be set of all function from $S$ to $F$.

It feels like an arbitrary choice, why $F^S$? Is there a deeper meaning or some historical reason?
Sorry if the question is stupid.
 A: I think your skepticism is very understandable! But here's the key fact that I think justifies that notation: if $F$ and $S$ are finite sets with $f$ and $s$ elements, respectively, then the number of functions from $S$ to $F$ (namely, the number of elements of $F^S$) equals $f^s$.
A: Bear with me....
Are you familiar with the concept of set foundation of the natural numbers?
It is the that $0= \emptyset$
$1 = \{0\} = \{\emptyset\}$.
$2 = \{0,1\} = \{\emptyset,\{\emptyset\}\}$
$3 = \{0,1,2\} = \{\emptyset,\{\emptyset\},\{\emptyset,\{\emptyset\}\}\}$ and so on.
I'm actually a little rusty on how exactly addition and muliplication of integers as sets are defined but... I trust it is consistently done and I could look it up if I wanted to.
Now in set theory everything is a set and a function from set $A$ to set $B$ is a subset of the set $A\times B= \{(a,b)|a\in A, b\in B\}$ which is a set of ordered pairs and ordered pairs are sets.  I confess I do not remember exactly how an ordered pair as an unordered set is defined or how a function as a set is defined. But again I trust is is done some way consistantly.
Now note.  If $n= \{0,1,2,3,...,n-1\}$ and $m=\{0,1,2,3,.....,m-1\}$ are sets.  Then if we consider the numbers for function $f_i:m \to n$ we realize that to count the number of functions we have for each $a \in \{0,1,2,....,m-1\}$ we have $f(a)\in \{0,1,2,3,...,n-1\}$ and we have $n$ choice for what $f(a)$ might be.  We consider and for another $b\in \{0,1,2,...,m-1\}; b\ne a$ we have $n$ choices of what $f(b)$ might be.  That is $n \times n = n^2$ choices for what $f(a),f(b)$ might be.  To count the total number of choices each of the $m$ values of $\{0,1,....,m-1\}$ are mapped to $n$ possible choices of $\{0,1,....,n-1\}$ so we have $\underbrace{n\times n \times n ....\times n}= n^m$ possible functions form $m \to n$.
That means we have a bijection from $\{f: f:m \to n\}$ and $n^m = \{0,1,2,.....,n^m-1\}$.  So $|\{f: f:m \to n\}| = |n^m| =n^m$.
Now as I said... I don't remember (actually I never really learned) the exact method in which the concept of function is defined in the set foundation model where "everything is a set" but I suspect (in fact idea be surprised if it weren't) that the actual set that represents the set of functions is exactly the same as the set $n^m$.
[for example  If we had $2= \{0,1\}$ and $3=\{0,1,2\}$ there are $3^2$ functions that mapp $2\to 3$.  The are $(0\to 0, 1\to 0),(0\to 0, 1\to 1),(0\to 0, 1\to 2),(0\to 1, 1\to 0),(0\to 1,1\to 1), (0\to 1, 1\to 2), (0\to 2,1\to 0),(0\to 2, 1\to 1),(0\to 2, 1\to 2)$.
[The set $3^2 = 9 = \{0,1,2,3,4,5,6,7,9\} = \{0\cdot 3+0,0\cdot 3+1, 0\cdot 3 + 2, 1\cdot 3+0, 1\cdot 3 + 1, 1\cdot 3 + 2, 2\cdot 3 + 0, 2\cdot 3 + 1, 2\cdot 3 + 2\}$.  So if we define the function $f_i:2\to 3 = (0\to a, 1\to b) = a\cdot 3 + b$ then we indeed do have that $\{f|f: 3\to 2\} = \{a\cdot 3 + b| a,b\in 3\}= \{0,...,8\}=9 = 3^2$ and that $3^2$ does equal the set of functions from $2\to 3$.]
[Now I suspect that isn't exactly how functions are defined but I suspect it is something quite analogous].
In any even, we can easily verify that $|\{f:S\to F\}| = |F|^{|S|}$ so at the very least it makes for a convenient shorthand notation.
As an aside you have probably seen the notation $|\mathbb R| = 2^{\mathbb N}$ meaning that $\mathbb R$ is uncountable whereas $|\mathbb Z| = |\mathbb N|$ means $\mathbb Z$ is countable.  This actually means that for any map for $\mathbb N \to \{0,1\}$ we which can be represented by a countably infinite string of zeros, and ones which can represent the binary expansion of a real number means the cardinality of the set of countably infinite binary strings is equal to the cardinality of the real numbers... which Cantor proved.
This notation seems odd at first but it makes sense if you think combinatorial about mapping elements from one set to another will result on $F^S$ possible mappings.
