Consequence of Nakayama Lemma's to Local Rings
Solution 1:
(Mariano's comment suffices but let me write the idea as an answer for the sake of completeness. This post is community wiki so I will not gain reputation for upvotes (or lose reputation for downvotes!).)
If $b_1,\dots,b_n\in M$ and if $\overline{b_1},\dots,\overline{b_n}$ is an $R/m$-basis for $M/mM$, then let $N=b_1A+\cdots+b_nA$ (be the submodule of $M$ generated by $b_1,\dots,b_n$).
Exercise 1: Prove that $N+mM=M$.
Exercise 2: Prove that $N=M$ by remembering that $R$ is a local ring and applying Nakayama's lemma.
I hope this helps!