[proofplan]
Work in the chosen triangularizing basis and write each action operator as an upper triangular matrix. The key computation is that the diagonal of a commutator of two upper triangular matrices is zero, because the diagonal entries of the two products agree. Since $[L,L]$ is spanned by Lie brackets and the representation sends Lie brackets to matrix commutators, every element of $[L,L]$ acts by a strictly upper triangular matrix. Finally, a strictly upper triangular $n \times n$ matrix has $n$th power zero.
[/proofplan]
[step:Write the representation in the triangularizing basis]
Let $n := \dim_F V$. If $n = 0$, then $V = 0$ and every endomorphism of $V$ is the zero endomorphism, so the conclusion is immediate. Assume henceforth that $n \geq 1$.
Let
\begin{align*}
A: L &\to M_n(F)
\end{align*}
denote the matrix representation of $\rho$ in the basis $(e_1,\dots,e_n)$; thus $A(x)$ is the matrix of the [linear map](/page/Linear%20Map) $\rho(x):V \to V$. By hypothesis, for every $x \in L$, the matrix $A(x)$ is upper triangular. Since $\rho$ is a Lie algebra representation, for all $x,y \in L$,
\begin{align*}
A([x,y]) = A(x)A(y) - A(y)A(x).
\end{align*}
[/step]
[step:Show that each basic commutator has zero diagonal]
Fix $x,y \in L$. Write $A := A(x)$ and $B := A(y)$. Both $A$ and $B$ are upper triangular matrices in $M_n(F)$.
For each index $i \in \{1,\dots,n\}$, the diagonal entry of $AB$ is
\begin{align*}
(AB)_{ii} = \sum_{k=1}^n A_{ik}B_{ki}.
\end{align*}
Because $A$ is upper triangular, $A_{ik}=0$ when $k<i$. Because $B$ is upper triangular, $B_{ki}=0$ when $k>i$. Hence every summand with $k \neq i$ is zero, and therefore
\begin{align*}
(AB)_{ii}=A_{ii}B_{ii}.
\end{align*}
The same argument gives
\begin{align*}
(BA)_{ii}=B_{ii}A_{ii}.
\end{align*}
Since $F$ is commutative,
\begin{align*}
(A([x,y]))_{ii}=(AB-BA)_{ii}=A_{ii}B_{ii}-B_{ii}A_{ii}=0.
\end{align*}
Also $AB-BA$ is upper triangular because products and differences of upper triangular matrices are upper triangular. Therefore $A([x,y])$ is strictly upper triangular.
[guided]
Fix $x,y \in L$ and abbreviate $A := A(x)$ and $B := A(y)$. The goal is to prove that the commutator matrix $AB-BA$ has zero diagonal. Since $A$ and $B$ are upper triangular, their products $AB$ and $BA$ are also upper triangular, so it remains only to compute the diagonal entries.
For an index $i \in \{1,\dots,n\}$, matrix multiplication gives
\begin{align*}
(AB)_{ii} = \sum_{k=1}^n A_{ik}B_{ki}.
\end{align*}
If $k<i$, then $A_{ik}=0$ because $A$ is upper triangular. If $k>i$, then $B_{ki}=0$ because $B$ is upper triangular. Thus the only possibly nonzero summand occurs at $k=i$, so
\begin{align*}
(AB)_{ii}=A_{ii}B_{ii}.
\end{align*}
Applying the same computation with $A$ and $B$ interchanged yields
\begin{align*}
(BA)_{ii}=B_{ii}A_{ii}.
\end{align*}
Since entries lie in the field $F$, multiplication of diagonal entries is commutative, so
\begin{align*}
(A([x,y]))_{ii}=(AB-BA)_{ii}=A_{ii}B_{ii}-B_{ii}A_{ii}=0.
\end{align*}
Thus every diagonal entry of $A([x,y])$ is zero. Since $AB-BA$ is still upper triangular, $A([x,y])$ is strictly upper triangular.
[/guided]
[/step]
[step:Extend the conclusion from brackets to the whole derived algebra]
By definition, $[L,L]$ is the $F$-linear span of all brackets $[x,y]$ with $x,y \in L$. Therefore, for any $z \in [L,L]$, there exist an integer $m \geq 1$, scalars $c_1,\dots,c_m \in F$, and elements $x_1,\dots,x_m,y_1,\dots,y_m \in L$ such that
\begin{align*}
z=\sum_{r=1}^m c_r[x_r,y_r].
\end{align*}
Using linearity of $A:L \to M_n(F)$,
\begin{align*}
A(z)=\sum_{r=1}^m c_r A([x_r,y_r]).
\end{align*}
Each matrix $A([x_r,y_r])$ is strictly upper triangular by the previous step, and the space of strictly upper triangular matrices is closed under finite $F$-linear combinations. Hence $A(z)$ is strictly upper triangular.
[/step]
[step:Deduce nilpotence from strict upper triangularity]
Let $z \in [L,L]$, and set $N := A(z) \in M_n(F)$. We have shown that $N$ is strictly upper triangular.
We prove that $N^n=0$. For each integer $q$ with $1 \leq q \leq n$, the matrix $N^q$ satisfies
\begin{align*}
(N^q)_{ij}=0 \quad \text{whenever } j-i<q.
\end{align*}
This is proved by induction on $q$. For $q=1$, it is exactly strict upper triangularity. If it holds for $q$, then
\begin{align*}
(N^{q+1})_{ij}=\sum_{k=1}^n (N^q)_{ik}N_{kj}.
\end{align*}
A summand can be nonzero only if $k-i \geq q$ and $j-k \geq 1$, hence only if $j-i \geq q+1$. Thus $(N^{q+1})_{ij}=0$ whenever $j-i<q+1$.
Taking $q=n$, every pair $i,j \in \{1,\dots,n\}$ satisfies $j-i \leq n-1<n$, so every entry of $N^n$ is zero. Therefore $N^n=0$, which means $\rho(z)^n=0$. Hence every element of $[L,L]$ acts nilpotently on $V$.
[/step]
