Series of a standard filiform Lie algebra

Put $U=\mathrm{span}(e_2,\dots ,e_n)$. Then $[U,U]=0$ and $[e_1,U]\subseteq U$. For a Jacobi triple entirely in $U$ every term is zero. For a triple containing exactly one copy of $e_1$, each possibly nonzero inner bracket lies in $U$ and is then bracketed with an element of $U$, so every term is zero. If a triple contains at least two copies of $e_1$, the two possibly nonzero terms cancel by bilinearity and the alternating law. Thus Jacobi holds, including in characteristic two, and the displayed rule defines a Lie algebra.

Every nonzero basis bracket is one of $e_3,\dots ,e_n$, and all of these occur. Hence ${\mathfrak{f}}_n^{(1)}=\mathrm{span}(e_3,\dots ,e_n)$. This subspace lies in the abelian space $U$, so ${\mathfrak{f}}_n^{(2)}=0$. Since $e_3\mathrm{\ne }0$ for $n\ge 3$, the least vanishing derived index is two by [L1].

The first bracket span gives ${\gamma }_2=\mathrm{span}(e_3,\dots ,e_n)$. If $2\le r\le n-2$ and ${\gamma }_r=\mathrm{span}(e_{r+1},\dots ,e_n)$, then only bracketing with $e_1$ contributes, and the displayed rule gives ${\gamma }_{r+1}=\mathrm{span}(e_{r+2},\dots ,e_n)$. Induction proves the promised formula through ${\gamma }_{n-1}=k{e}_n\mathrm{\ne }0$; finally $[{\mathfrak{f}}_n,k{e}_n]=0$, so ${\gamma }_n=0$. By [L2] and [L3], the class is exactly $n-1$.

When $n=3$, the calculation reads ${\gamma }_2=k{e}_3\mathrm{\ne }0$ and ${\gamma }_3=0$, so the smallest permitted dimension has class two and derived length two. For every $n\ge 3$, steps 2.1 and 2.2 establish both exact endpoints using only the displayed finite basis; no choice principle is used.