An algebra $A$ is called nilpotent if $A^n=0$ for some positive integer $n$. Also, we call algebra $A$ is solvable if $A^{(n)}=0$, with solvable index $n$, where $A^{(n)} = A^{(n-1)}A^{(n-1)}$.
Can we say that if algebra $A$ is solvable with solvable index $n$, then $A$ is nilpotent with nilptency index $n$ or $k < n$?. In other words is it true that $A^{n}\subseteq A^{(n)}$?
This isn’t true even for Lie algebras, take for example the 2 dimensional non Abelian Lie algebra, it is solvable but not nilpotent