= Solution
The <endomorphisms> of the <finite-dimensional vector space> $V$ over the <complex numbers> form the <general linear Lie algebra> $L=\operatorname{End}_{\mathbb C}(V)$ with the usual addition and scalar multiplication and <Lie bracket> $[x,y]=xy-yx$. This <commutator> is bilinear and antisymmetric, and expanding the six products verifies the <Jacobi identity>.
For a <Lie subalgebra> $\mathfrak g=L_1$, being an <abelian Lie algebra> means $[\mathfrak g,\mathfrak g]=0$. A <nilpotent Lie algebra> has $\gamma_1=\mathfrak g$, $\gamma_{j+1}=[\mathfrak g,\gamma_j]$ eventually zero; a <solvable Lie algebra> has $\mathfrak g^{(0)}=\mathfrak g$, $\mathfrak g^{(j+1)}=[\mathfrak g^{(j)},\mathfrak g^{(j)}]$ eventually zero. These are the <lower central series of a Lie algebra> and <derived series of a Lie algebra>, respectively. Nilpotence is a condition on the <Lie bracket>, and does not require every member to be a <nilpotent endomorphism>: a nonzero scalar multiple of the identity spans an <abelian Lie algebra>.
A <flag of a vector space> is an increasing chain of <vector subspaces>. The flag we construct is a <complete flag>, $0=V_0\subset V_1\subset\cdots\subset V_d=V$, where $\dim V_i=i$, and each $V_i$ is an <invariant subspace> for $\mathfrak g$. We first prove the common-<eigenvector> assertion in the <Lie theorem>, by induction on $\dim\mathfrak g$; the zero algebra is immediate. For nonzero solvable $\mathfrak g$, its <derived algebra> is proper, so there is a codimension-one <ideal of a Lie algebra> $\mathfrak k$ containing $[\mathfrak g,\mathfrak g]$. Write $\mathfrak g=\mathfrak k\oplus\mathbb Cx$. By induction there are $v\ne0$ and a <linear functional> $\lambda$ on $\mathfrak k$ such that $kv=\lambda(k)v$ for every $k\in\mathfrak k$.
Let $W$ be the <cyclic subspace> spanned by $v,xv,x^2v,\ldots$. The <commutator derivation identity> and $[\mathfrak k,x]\subseteq\mathfrak k$ show inductively that
$$
kx^jv\equiv\lambda(k)x^jv\pmod{\operatorname{span}\{v,xv,\ldots,x^{j-1}v\}}.
$$
Thus $W$ and all its initial cyclic spans are $\mathfrak k$-invariant. If $m=\dim W$, the first $m$ cyclic vectors form a <basis>, $W$ is also $x$-invariant, and $\operatorname{tr}_W k=m\lambda(k)$. For $k\in\mathfrak k$, the <trace of a matrix commutator> gives
$$
0=\operatorname{tr}_W[k,x]=m\lambda([k,x]).
$$
Hence $\lambda([k,x])=0$, since the field has <characteristic zero>. The nonzero common <weight space>
$$
V_\lambda=\{w\in V:kw=\lambda(k)w\text{ for all }k\in\mathfrak k\}
$$
is $x$-invariant: $k(xw)=x(kw)+[k,x]w=\lambda(k)xw$. The restriction of $x$ to $V_\lambda$ has an <eigenvector>, because $\mathbb C$ is an <algebraically closed field>. This is a common <eigenvector> for $\mathfrak g$. Its line is invariant, and repeating the argument on the <quotient vector space> gives the complete invariant flag. Equivalently, this proves <simultaneous triangularization of a Lie algebra representation>.
In a <basis> adapted to this <complete flag>, every member of $\mathfrak g$ is upper triangular, so every member of its <derived algebra> is strictly upper triangular. Products of $d$ <strictly upper triangular matrices> vanish, and each iterated <Lie bracket> of $d$ such matrices is a sum of these products. Consequently the <derived algebra> is a <nilpotent Lie algebra>. We may therefore take \b[$L_2=[L_1,L_1]$]: it is an ideal, and the <quotient Lie algebra> $L_1/L_2$ is abelian. This also covers $\dim V=0$.
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