Nilpotent Lie algebra

Remarks

Equivalently, there is $n$ such that $[x_1, [x_2, [\cdots [x_{n-1}, x_n] \cdots]]] = 0$ for all $x_1, \dots, x_n \in \mathfrak g$. Abelian Lie algebras are nilpotent, with $\mathfrak g^2 = 0$. By induction, the derived series satisfies $\mathfrak g^{(i)} \subseteq \mathfrak g^{i+1}$, so a nilpotent Lie algebra is solvable.

Example: the Heisenberg algebra with basis $x, y, z$, $[x, y] = z$ and $z$ central. Here $\mathfrak g^2 = kz$ and $\mathfrak g^3 = 0$.

Non-example: the two-dimensional Lie algebra with basis $x, y$ and $[x, y] = y$. Here $\mathfrak g^2 = ky$ and $\mathfrak g^3 = [\mathfrak g, ky] = ky$ because $[x, y] = y$, so $\mathfrak g^i = ky$ for all $i \ge 2$. This algebra is solvable but not nilpotent.