If S is a nonempty set, then $F^S$ is a vector space over F I was given a task to prove the following statement

If $S$ is a nonempty set, then $F^S$ is a vector space over $F$ .

Definitions:

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

I was trying to construct a proof, but do not even know where to start. Could you please suggest me any idea how can I prove this statement ?
I will try to construct proof based on the advices in the comments:
There are 2 operations on $F^N$:
addition $F^S : (f + g) (x) = f(x) + g(x)$
multiplication $F^S : (zf) (x) = zf(x)$
In order for a set V to be a vector space operations of addition and multiplication should be defined ( we already have it ) and following properties should held:
associativity
commutativity
distributivity
additive inverse
additive identity

Hint: You only need to verify the vector space axioms. For example, you need to show that $F^S$ is an abelian group with the sum of vectors. This sum is defined as follows: Given two vectors $f,g\in F^S$ i,e, $f,g:S\rightarrow F$, define $(f+g)(s)=f(s)+g(s)$ for every $s\in S$ (remember that $F$ must be a field). Note that this sum satisfy  $f+g=g+f$, because for every $s\in S$ $(f+g)(s)=f(s)+g(s)=g(s)+f(s)=(g+f)(s)$ and in $F$ the sum is also conmutative. In the same way you can proof the other axioms of abelian group. The axioms realted with scalars are also easy to check, you only need to use the right definition of product between scalars and vectors.
