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Consider a given set of structure constants $c_{ij}^k$ defining a (finite dimensional) Lie algebra $\mathfrak{L}$, i.e. $$[e_i,e_j] = \sum_{k=1}^N c_{i,j}^k \, e_k \qquad i,j=1,\ldots,N$$ with $N$ denoting the dimension of $\mathfrak{L}$. In terms of the $c_{i,j}^k$ the adjoint representation of the basis element $e_i$ is the matrix $$ ad(e_i) \; = \; \begin{pmatrix}c_{i,1}^1 & \ldots & c_{i,N}^1\\ \vdots & & \vdots\\ c_{i,1}^N & \ldots & c_{i,N}^N\end{pmatrix} \qquad i=1,\ldots,N $$ Assuming the sequence $$[ad(e_1),[ad(e_2),[\ldots[ad(e_{N-1}),ad(e_N)]]]] = 0$$ vanishes, is it correct to conclude that $\mathfrak{L}$ is nilpotent? (Wikipedia describes the required length of the sequence as "sufficiently large". Is $N$ sufficiently large?)

user71769
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    You need more N-fold commutators to vanish, namely, all N-fold commutators where we are using all orders of generators and even using some generators repeatedly. Just think of commutative case. With this in mind, the proof us a simple induction on N. – Moishe Kohan Feb 01 '14 at 22:13

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