On some notes, I have found on Internet, states the following argument, without proof
If $A$ is a free abelian group with $r(A)=n$ and $A=\langle s_1, \dots , s_n \rangle$ then $S=\{s_1, \dots s_n\}$ is a basis of $A$.
In general, if $A$ is abelian group then $S=\{s_1, \dots s_n\} \subseteq A$ is a basis of $A$ if
$$\forall a \in A, \,\exists !\, \lambda_i \in \mathbb{Z} : a=\lambda_1 s_1+\cdots+\lambda_ns_n$$
or equivalently
$$A=\langle S \rangle \text{ and } S \text{ is } \mathbb{Z}\text{-linear independent set}$$
So, i tried to prove by myself the argument but I have not a clue how to proceed. Some help would be appreciated.
p.s. I have a feeling that the answer is ridiculously simple but a i have really stacked