[proofplan] We filter the upper unitriangular group by how far above the diagonal its first possible nonzero entries may occur. The key estimate is that a commutator of an element supported from the $r$th superdiagonal onward with an element supported from the $s$th superdiagonal onward is supported from the $(r+s)$th superdiagonal onward. Applying this estimate inductively to the lower central series forces $\gamma_k$ into the $k$th filtration subgroup, and the $n$th filtration subgroup is the identity. [/proofplan] [step:Define the superdiagonal filtration] Let $G:=U_n^{\mathrm{up}}(\mathbb R)$. For an integer $r\geq 1$, define $E_r$ to be the set of all real $n\times n$ matrices $X=(X_{ij})$ such that \begin{align*}X_{ij}=0 \quad \text{whenever} \quad j-i<r.\end{align*} Thus $E_r$ consists of matrices whose possible nonzero entries lie on the $r$th superdiagonal or higher. If $r\geq n$, then $E_r=\{0\}$, because there is no pair $(i,j)$ with $1\leq i,j\leq n$ and $j-i\geq n$. For $r\geq 1$, define \begin{align*}F_r:=\{I+X:X\in E_r\}.\end{align*} Since every $X\in E_1$ is strictly upper triangular, $F_1=G$. Since $E_n=\{0\}$, we have $F_n=\{I\}$. We also record that each $F_r$ is a subgroup of $G$. If $X,Y\in E_r$, then $X+Y+XY\in E_r$, because $E_r$ is closed under addition and multiplication by an upper triangular matrix cannot create entries below the $r$th superdiagonal. Hence \begin{align*}(I+X)(I+Y)=I+(X+Y+XY)\in F_r.\end{align*} Also $X$ is strictly upper triangular, so $X^n=0$, and therefore \begin{align*}(I+X)^{-1}=I-X+X^2-\cdots+(-1)^{n-1}X^{n-1}.\end{align*} The nonidentity part of this finite sum lies in $E_r$, so $(I+X)^{-1}\in F_r$. Thus $F_r\leq G$. [guided] The filtration is designed to measure how far away from the diagonal the first possible nonzero entries are. For $r\geq 1$, the condition $X\in E_r$ means that the entries on the diagonal and on the first $r-1$ superdiagonals vanish. In coordinates this is exactly \begin{align*}X_{ij}=0 \quad \text{whenever} \quad j-i<r.\end{align*} The number $j-i$ is the superdiagonal height of the entry $(i,j)$: height $0$ is the diagonal, height $1$ is the first superdiagonal, and so on. We then define \begin{align*}F_r:=\{I+X:X\in E_r\}.\end{align*} For $r=1$, the condition says that $X$ vanishes whenever $j-i<1$, so $X$ has only strictly upper triangular entries. Thus $I+X$ is precisely an upper triangular matrix with all diagonal entries equal to $1$, and $F_1=G$. For $r=n$, no matrix entry can satisfy $j-i\geq n$, since $1\leq i,j\leq n$ implies $j-i\leq n-1$. Hence $E_n=\{0\}$ and $F_n=\{I\}$. We need these $F_r$ to be subgroups, not just subsets, because the lower central series is defined using subgroups generated by commutators. Let $X,Y\in E_r$. Then \begin{align*} (I+X)(I+Y)=I+(X+Y+XY). \end{align*} The matrices $X$ and $Y$ have no entries below height $r$. The product $XY$ also has no entries below height $r$: multiplying on the right by an upper triangular matrix cannot move a nonzero entry closer to the diagonal. Therefore $X+Y+XY\in E_r$, and the product lies in $F_r$. For inverses, use the fact that every strictly upper triangular $n\times n$ matrix is nilpotent of index at most $n$. Thus $X^n=0$, and the finite geometric series gives \begin{align*} (I+X)^{-1}=I-X+X^2-\cdots+(-1)^{n-1}X^{n-1}. \end{align*} Each power $X^m$ with $m\geq 1$ still has no entries below the $r$th superdiagonal, so the nonidentity part of the inverse belongs to $E_r$. Hence $(I+X)^{-1}\in F_r$. This proves $F_r\leq G$. [/guided] [/step] [step:Show products add superdiagonal heights] We claim that if $r,s\geq 1$, $X\in E_r$, and $Y\in E_s$, then $XY\in E_{r+s}$. Indeed, for $1\leq i,j\leq n$, \begin{align*}(XY)_{ij}=\sum_{\ell=1}^{n}X_{i\ell}Y_{\ell j}.\end{align*} If $j-i<r+s$, then for each $\ell$ either $\ell-i<r$ or $j-\ell<s$; otherwise adding the inequalities $\ell-i\geq r$ and $j-\ell\geq s$ would give $j-i\geq r+s$. Hence each summand $X_{i\ell}Y_{\ell j}$ is zero, so $(XY)_{ij}=0$. Therefore $XY\in E_{r+s}$. The same argument with $X$ and $Y$ interchanged gives $YX\in E_{r+s}$. [/step] [step:Prove that commutators move into the summed filtration level] Let $r,s\geq 1$, let $A\in F_r$, and let $B\in F_s$. Choose $X\in E_r$ and $Y\in E_s$ such that \begin{align*}A=I+X, \qquad B=I+Y.\end{align*} By the previous step, $XY\in E_{r+s}$ and $YX\in E_{r+s}$, so \begin{align*}AB-BA=(I+X)(I+Y)-(I+Y)(I+X)=XY-YX\in E_{r+s}.\end{align*} Since $A^{-1}$ and $B^{-1}$ are upper triangular, their product $A^{-1}B^{-1}$ is upper triangular. Multiplying a matrix in $E_{r+s}$ on the right by an upper triangular matrix preserves membership in $E_{r+s}$. Therefore \begin{align*}ABA^{-1}B^{-1}-I=(AB-BA)A^{-1}B^{-1}\in E_{r+s}.\end{align*} Thus every commutator $ABA^{-1}B^{-1}$ with $A\in F_r$ and $B\in F_s$ lies in $F_{r+s}$. Since $F_{r+s}$ is a subgroup, the subgroup generated by all such commutators is contained in $F_{r+s}$. Hence \begin{align*} [F_r,F_s]\subset F_{r+s}. \end{align*} [guided] The goal is to prove that commutators increase superdiagonal height. Take $A\in F_r$ and $B\in F_s$. By definition of the filtration, there are matrices $X\in E_r$ and $Y\in E_s$ with \begin{align*}A=I+X, \qquad B=I+Y.\end{align*} The preceding step says that products add heights: $XY\in E_{r+s}$ and $YX\in E_{r+s}$. Now compute the failure of $A$ and $B$ to commute: \begin{align*}AB-BA=(I+X)(I+Y)-(I+Y)(I+X)=XY-YX.\end{align*} Both $XY$ and $YX$ lie in $E_{r+s}$, and $E_{r+s}$ is closed under subtraction, so \begin{align*}AB-BA\in E_{r+s}.\end{align*} The group commutator is not $AB-BA$; it is \begin{align*}ABA^{-1}B^{-1}.\end{align*} To connect these two expressions, subtract the identity and factor: \begin{align*}ABA^{-1}B^{-1}-I=ABA^{-1}B^{-1}-BAA^{-1}B^{-1}=(AB-BA)A^{-1}B^{-1}.\end{align*} This identity is the reason the additive estimate on $AB-BA$ controls the multiplicative commutator. It remains to check that multiplying on the right by $A^{-1}B^{-1}$ does not create lower superdiagonal entries. Since $A$ and $B$ are upper triangular unipotent matrices, their inverses are upper triangular, and therefore $A^{-1}B^{-1}$ is upper triangular. If $Z\in E_{r+s}$ and $T$ is upper triangular, then in \begin{align*} (ZT)_{ij}=\sum_{\ell=1}^{n}Z_{i\ell}T_{\ell j}, \end{align*} a nonzero summand requires $\ell-i\geq r+s$ and $j-\ell\geq 0$, hence $j-i\geq r+s$. Thus $ZT\in E_{r+s}$. Applying this with $Z=AB-BA$ and $T=A^{-1}B^{-1}$ gives \begin{align*} ABA^{-1}B^{-1}-I\in E_{r+s}. \end{align*} Therefore $ABA^{-1}B^{-1}\in F_{r+s}$. Finally, $[F_r,F_s]$ denotes the subgroup generated by all commutators $ABA^{-1}B^{-1}$ with $A\in F_r$ and $B\in F_s$. Since every such commutator lies in the subgroup $F_{r+s}$, the whole generated subgroup lies in $F_{r+s}$. Hence \begin{align*} [F_r,F_s]\subset F_{r+s}. \end{align*} [/guided] [/step] [step:Apply the commutator estimate to the lower central series] As in the statement, define the lower central series of $G$ by $\gamma_1(G):=G$ and $\gamma_{k+1}(G):=[\gamma_k(G),G]$ for every integer $k\geq 1$, where $[H,K]$ denotes the subgroup generated by all commutators $hkh^{-1}k^{-1}$ with $h\in H$ and $k\in K$. We prove by induction that \begin{align*}\gamma_k(G)\subset F_k\end{align*} for every $k\geq 1$. For $k=1$, this is $\gamma_1(G)=G=F_1$. Suppose $\gamma_k(G)\subset F_k$. Since $G=F_1$, every commutator with first entry in $\gamma_k(G)$ and second entry in $G$ is also a commutator with first entry in $F_k$ and second entry in $F_1$. Therefore the subgroup generated by the former commutators is contained in the subgroup generated by the latter commutators. Using the commutator estimate, we obtain \begin{align*}\gamma_{k+1}(G)=[\gamma_k(G),G]\subset [F_k,F_1]\subset F_{k+1}.\end{align*} This completes the induction. [/step] [step:Conclude nilpotence from the terminal filtration subgroup] Taking $k=n$ in the induction result gives \begin{align*} \gamma_n(G)\subset F_n=\{I\}. \end{align*} Since every term of the lower central series is a subgroup of $G$, this implies \begin{align*} \gamma_n(G)=\{I\}. \end{align*} Therefore the lower central series reaches the identity after at most $n-1$ nontrivial commutator steps. Hence $U_n^{\mathrm{up}}(\mathbb R)$ is nilpotent of nilpotency class at most $n-1$. [/step]