As one says, proofs are useful not only to certify the truth of a statement, but also to remember and understand the statement better. But in this second function, proofs should not be understood only as "proof of the statement". Proofs that use the statement, instead, are also a very useful way to remember it.

There was a time where I had difficulty to remember Nakayama. This time ended when I learnt the following basic result: Let $M$ be a finitely generated module over a local domain $A$ of maximal ideal $m$, residue field $A/m=k$, fraction field $K$. To such an $M$ one can attach two finite-dimensional vector spaces, one, $M \otimes_A K$ over $K$, the other, $M \otimes_A k = M/mM$ over $k$. Then one always $\dim_K M \otimes_A K \leq \dim_k M \otimes_A k$ with equality if and only if $M$ is free.

I find this result much more striking and easy to remember than Nakayama itself. Yet it is essentially equivalent to it. Here is the proof: Take $e_1, \dots, e_n$ be a basis of $M \otimes_A k$. Lift this in a family $f_1,\dots,f_n$ of $A$. By Nakayama, $f_1, \dots, f_n$ generates $M$ as an $A$-module, hence $M \otimes_A K$ as a $K$-vector space, hence the stated inequality. If furthermore is is an equality, then $f_1,\dots,f_n$ is a basis of $M \otimes_A K$, hence $K$-free, hence $A$-free, hence an $A$-basis of $M$.

As one says, proofs are useful not only to certify the truth of a statement, but also to remember and understand the statement better. But in this second function, proofs should not be understood only as "proof of the statement". Proofs that use the statement, instead, are also a very useful way to remember it.

There was a time where I had difficulty to remember Nakayama. This time ended when I learnt the following basic result: Let $M$ be a finitely generated module over a local domain $A$ of maximal ideal $m$, residue field $A/m=k$, fraction field $K$. To such an $M$ one can attach two finite-dimensional vector spaces, one, $M \otimes_A K$ over $K$, the other, $M \otimes_A k = M/mM$ over $k$. Then one always has $\dim_K M \otimes_A K \leq \dim_k M \otimes_A k$ with equality if and only if $M$ is free.

I find this result much more striking and easy to remember than Nakayama itself. Yet it is essentially equivalent to it. Here is the proof: Take $e_1, \dotsc, e_n$ be a basis of $M \otimes_A k$. Lift this in a family $f_1,\dotsc,f_n$ of $M$. By Nakayama, $f_1, \dotsc, f_n$ generates $M$ as an $A$-module, hence $M \otimes_A K$ as a $K$-vector space, hence the stated inequality. If furthermore it is an equality, then $f_1,\dotsc,f_n$ is a basis of $M \otimes_A K$, hence $K$-free, hence $A$-free, hence an $A$-basis of $M$.