On Lie nilpotent associative algebras

Let $G$ be a group generated by a set $X$. It is well known and easy to check that $$[g_1,g_2,\dots ,g_n]=1forall{g}_i\in G⟺[x_1,x_2,\dots ,x_n]=1forall{x}_i\in X\mathrm{.}$$ Let $L$ be a Lie algebra generated by a set $X$. Then it is also well known and easy to check that $$[h_1,h_2,\dots ,h_n]=0forall{h}_i\in L⟺[x_1,x_2,\dots ,x_n]=0forall{x}_i\in X\mathrm{.}$$ Now let $A$ be a unital associative algebra generated by a set $X$. Then the assertion similar to the above does not hold: for $n>2$, it is easy to find an algebra $A$ with a generating set $X$ such that $[x_1,x_2,\dots ,x_n]=0$ for all $x_i\in X$ but $[a_1,a_2,\dots ,a_n]\mathrm{\ne }0$ for some $a_i\in A$. However, we prove the following result. Let $R$ be a unital associative and commutative ring such that $\frac{1}{3}\in R$. Let $A$ be a unital associative $R$-algebra generated by a set $X$. Let $X^2=\{x_1{x}_2\mid x_i\in X\}$ be the set of all products of $2$ elements of $X$. Then $$[a_1,a_2,\dots ,a_n]=0forall{a}_i\in A⟺[y_1,y_2,\dots ,y_n]=0forall{y}_i\in X\cup X^2\mathrm{.}$$ Moreover, one can assume that in the commutator $[y_1,y_2,\dots ,y_n]$ above $y_1,y_n\in X$.