[proofplan] We show that the first derived algebra of $\mathfrak b_n(F)$ consists of strictly upper triangular matrices by computing diagonal entries of commutators. We then introduce a filtration of the strictly upper triangular matrices by vanishing of successive superdiagonals. Matrix multiplication adds filtration degrees, so taking brackets inside the filtration moves matrices farther above the diagonal. The derived series therefore enters filtration levels beyond the last possible superdiagonal and becomes zero. [/proofplan]

[step:Show commutators of upper triangular matrices have zero diagonal]Let $\mathfrak b := \mathfrak b_n(F)$. For $A, B \in \mathfrak b$, write $A_{ij} \in F$ and $B_{ij} \in F$ for their matrix entries. Since $A$ and $B$ are upper triangular, $A_{ik}=0$ and $B_{ik}=0$ whenever $k<i$. For each $i \in \{1,\dots,n\}$, the $i$th diagonal entry of $AB$ is \begin{align*} (AB)_{ii} &= \sum_{k=1}^{n} A_{ik}B_{ki}. \end{align*} If $k<i$, then $A_{ik}=0$; if $k>i$, then $B_{ki}=0$. Hence only the term $k=i$ can contribute, so \begin{align*} (AB)_{ii} = A_{ii}B_{ii}. \end{align*} The same argument gives \begin{align*} (BA)_{ii} = B_{ii}A_{ii}. \end{align*} Because $F$ is commutative, \begin{align*} [A,B]_{ii} &= (AB-BA)_{ii} \\ &= A_{ii}B_{ii}-B_{ii}A_{ii} \\ &= 0. \end{align*} Since $AB$ and $BA$ are upper triangular, $[A,B]$ is upper triangular with zero diagonal. Therefore \begin{align*} [\mathfrak b,\mathfrak b]\subseteq \mathfrak n_{n,1}(F), \end{align*} where $\mathfrak n_{n,1}(F)$ denotes the strictly upper triangular matrices in $M_n(F)$.[/step]

[guided]The first goal is to identify where commutators of upper triangular matrices can live. Let $\mathfrak b := \mathfrak b_n(F)$, and take arbitrary matrices $A,B \in \mathfrak b$. For entries, write $A_{ij},B_{ij}\in F$. Upper triangularity means $A_{ij}=B_{ij}=0$ whenever $i>j$. Fix a diagonal position $i\in\{1,\dots,n\}$. The diagonal entry of the product $AB$ is computed by the definition of matrix multiplication: \begin{align*} (AB)_{ii} &= \sum_{k=1}^{n} A_{ik}B_{ki}. \end{align*} For $k<i$, the entry $A_{ik}$ lies below the diagonal of $A$, so $A_{ik}=0$. For $k>i$, the entry $B_{ki}$ lies below the diagonal of $B$, so $B_{ki}=0$. Thus every term except $k=i$ vanishes, and \begin{align*} (AB)_{ii}=A_{ii}B_{ii}. \end{align*} Repeating the same computation with the product $BA$ gives \begin{align*} (BA)_{ii}=B_{ii}A_{ii}. \end{align*} Since $F$ is a field, its multiplication is commutative, so $A_{ii}B_{ii}=B_{ii}A_{ii}$. Therefore \begin{align*} [A,B]_{ii} &=(AB-BA)_{ii}\\ &=A_{ii}B_{ii}-B_{ii}A_{ii}\\ &=0. \end{align*} The product of upper triangular matrices is upper triangular, so $AB$ and $BA$ are upper triangular, and hence $AB-BA$ is also upper triangular. We have proved that every commutator in $\mathfrak b$ is strictly upper triangular. Therefore \begin{align*} [\mathfrak b,\mathfrak b]\subseteq \mathfrak n_{n,1}(F), \end{align*} where $\mathfrak n_{n,1}(F)$ is the subspace of strictly upper triangular matrices.[/guided]

[step:Filter strictly upper triangular matrices by superdiagonal order]For each integer $r\ge 1$, define the $F$-linear subspace \begin{align*} \mathfrak n_{n,r}(F) := \{A\in M_n(F): A_{ij}=0 \text{ whenever } j-i<r\}. \end{align*} Thus $\mathfrak n_{n,1}(F)$ is the strictly upper triangular subspace. Also, \begin{align*} \mathfrak n_{n,r}(F)=0 \end{align*} for every $r\ge n$, because every pair of indices $1\le i,j\le n$ satisfies $j-i\le n-1$. We claim that for all integers $r,s\ge 1$, \begin{align*} \mathfrak n_{n,r}(F)\,\mathfrak n_{n,s}(F)\subseteq \mathfrak n_{n,r+s}(F). \end{align*} Indeed, let $A\in \mathfrak n_{n,r}(F)$ and $B\in \mathfrak n_{n,s}(F)$. Fix $i,j\in\{1,\dots,n\}$ with $j-i<r+s$. For each $k\in\{1,\dots,n\}$, if $k-i<r$, then $A_{ik}=0$. If $k-i\ge r$, then \begin{align*} j-k=(j-i)-(k-i)<(r+s)-r=s, \end{align*} so $B_{kj}=0$. Hence every summand in \begin{align*} (AB)_{ij}=\sum_{k=1}^{n}A_{ik}B_{kj} \end{align*} is zero. Thus $(AB)_{ij}=0$ whenever $j-i<r+s$, proving $AB\in \mathfrak n_{n,r+s}(F)$. Consequently, \begin{align*} [\mathfrak n_{n,r}(F),\mathfrak n_{n,s}(F)] \subseteq \mathfrak n_{n,r+s}(F) \end{align*} for all $r,s\ge 1$, because both products $AB$ and $BA$ lie in $\mathfrak n_{n,r+s}(F)$.[/step]

[guided]We now measure how far above the diagonal a matrix begins. For each integer $r\ge 1$, define \begin{align*} \mathfrak n_{n,r}(F) := \{A\in M_n(F): A_{ij}=0 \text{ whenever } j-i<r\}. \end{align*} The condition $j-i<1$ is the condition $j\le i$, so $\mathfrak n_{n,1}(F)$ consists exactly of strictly upper triangular matrices. Larger values of $r$ force more superdiagonals to vanish. Since the largest possible value of $j-i$ in an $n\times n$ matrix is $n-1$, we have \begin{align*} \mathfrak n_{n,r}(F)=0 \end{align*} for all $r\ge n$. The key algebraic fact is that multiplying matrices adds these filtration levels. Let $r,s\ge 1$, let $A\in \mathfrak n_{n,r}(F)$, and let $B\in \mathfrak n_{n,s}(F)$. We prove \begin{align*} AB\in \mathfrak n_{n,r+s}(F). \end{align*} Fix indices $i,j\in\{1,\dots,n\}$ satisfying $j-i<r+s$. We must show $(AB)_{ij}=0$. By matrix multiplication, \begin{align*} (AB)_{ij}=\sum_{k=1}^{n}A_{ik}B_{kj}. \end{align*} For a fixed summation index $k$, there are two cases. If $k-i<r$, then $A_{ik}=0$ by the definition of $\mathfrak n_{n,r}(F)$. If $k-i\ge r$, then \begin{align*} j-k=(j-i)-(k-i)<(r+s)-r=s, \end{align*} so $B_{kj}=0$ by the definition of $\mathfrak n_{n,s}(F)$. Thus in every case the product $A_{ik}B_{kj}$ is zero. Therefore the whole sum is zero: \begin{align*} (AB)_{ij}=0. \end{align*} Since this holds for every pair $(i,j)$ with $j-i<r+s$, we have $AB\in \mathfrak n_{n,r+s}(F)$. Applying the same argument to $BA$ shows $BA\in \mathfrak n_{n,r+s}(F)$ as well. Because $\mathfrak n_{n,r+s}(F)$ is an $F$-linear subspace, the difference $AB-BA$ also lies in it. Hence \begin{align*} [\mathfrak n_{n,r}(F),\mathfrak n_{n,s}(F)] \subseteq \mathfrak n_{n,r+s}(F). \end{align*} This is the mechanism that makes repeated commutators move upward through the superdiagonals.[/guided]

[step:Bound the derived series by the filtration] Define the derived series of $\mathfrak b$ by \begin{align*} \mathfrak b^{(0)}&:=\mathfrak b,\\ \mathfrak b^{(m+1)}&:=[\mathfrak b^{(m)},\mathfrak b^{(m)}]\quad\text{for }m\ge 0. \end{align*} We prove by induction on $m\ge 1$ that \begin{align*} \mathfrak b^{(m)}\subseteq \mathfrak n_{n,2^{m-1}}(F). \end{align*} For $m=1$, the first step gives \begin{align*} \mathfrak b^{(1)}=[\mathfrak b,\mathfrak b]\subseteq \mathfrak n_{n,1}(F) =\mathfrak n_{n,2^{0}}(F). \end{align*} Assume that for some $m\ge 1$, \begin{align*} \mathfrak b^{(m)}\subseteq \mathfrak n_{n,2^{m-1}}(F). \end{align*} Using the bracket filtration estimate with $r=s=2^{m-1}$, we obtain \begin{align*} \mathfrak b^{(m+1)} &=[\mathfrak b^{(m)},\mathfrak b^{(m)}]\\ &\subseteq [\mathfrak n_{n,2^{m-1}}(F),\mathfrak n_{n,2^{m-1}}(F)]\\ &\subseteq \mathfrak n_{n,2^m}(F). \end{align*} This completes the induction. Choose $m\ge 1$ such that $2^{m-1}\ge n$. Then \begin{align*} \mathfrak b^{(m)} \subseteq \mathfrak n_{n,2^{m-1}}(F) =0. \end{align*} Thus the derived series of $\mathfrak b_n(F)$ terminates at zero, so $\mathfrak b_n(F)$ is solvable. [/step]