I am stuck with the following: let $(A,\mathfrak{m},k)$ be a commutative local ring with unity, and $F$ a free $A$-module of rank $n$. Then: $$\{e_i\}_{i=1}^n \text{ is a basis for } F \text{ as an $A$-module} \iff \{e_i+\mathfrak{m}F \}_{i=1}^n \text{ is a basis for } F/\mathfrak{m}F \text{ as a $k$-vector space}$$
What I tried so far: I know that, as a consequence of Nakayama's Lemma, the set on the left generates $F$ as an $A$-module if, and only if, the set on the right generates $F/\mathfrak{m}F$ as a $k$-vector space. If the $e_i$ form a basis, and the $e_i+\mathfrak{m}F$ don't, then we can extract a basis $\{e_{i_j}+\mathfrak{m}F\}_{j=1}^m$, $m<n$ (we are in a vector space), and then $\{e_{i_j}\}_{j=1}^m$ would generate $F$, which can't happen. For the converse, I'm a bit lost, as if the $e_i+\mathfrak{m}F$ form a basis, the $e_i$ form a generating set for the free module $F$ of rank $n$, and I know that this implies they are in fact a basis, but the proofs of this fact that I found use tensor products or other more advanced tools, and the book only covered at this point basic facts about modules and localizations. Any ideas?
Let $F$ be your free $A$-module, and let $A^n = Ab_1 \oplus \ldots \oplus Ab_n$ (we could use $F$ here, but this is confusing).
Define a map $f: A^n \to F$ by $f(b_i) = e_i$. By assumption, the $e_i$ generate $F$ so this map $f$ is surjective. Since $F$ is free, hence projective, there exists a map $g: F \to A^n$ such that $fg$ is the identity and $A^n = F \oplus \ker f$.
Now, if $\sum r_ib_i \in \ker f$, then $\sum r_i e_i = 0$, hence $r_i \in \mathfrak m$ for all $i$ since $r_i + \mathfrak m = 0$ for all $i$.
Thus, $\ker f \subset \mathfrak mA^n = \mathfrak m(F \oplus \ker f) = \mathfrak mF \oplus \mathfrak m\ker f$, so $\mathfrak m \ker f = \ker f$. We have that $\ker f$ is finitely generated being a direct summand of $A^n$, so by Nakayama's Lemma $\ker f = 0$ (here we use that $A$ is local). Thus $f$ is an isomorphism and the $e_i$ form a basis.
More generally this works for any projective $F$, and even when $F$ is not finitely generated. For a proof of this, you could take a look at Matsumura's Commutative Ring Theory, p. 10.