[proofplan] We prove the displayed formula by induction on the lower central series index. The base case is the definition of the quotient Lie algebra. The induction step uses the fact that taking Lie brackets in the quotient sends representatives $x+I$ and $y+I$ to $[x,y]+I$, so the bracket of $\mathfrak g/I$ with $(\gamma_i(\mathfrak g)+I)/I$ is exactly $([\mathfrak g,\gamma_i(\mathfrak g)]+I)/I$. Once the formula is established, nilpotence descends because a vanishing term $\gamma_{c+1}(\mathfrak g)=0$ maps to the zero subspace of the quotient. [/proofplan]

[step:Define the bracket notation and the quotient bracket] For $k$-linear subspaces $A,B \subseteq \mathfrak g$, define \begin{align*} [A,B] := \operatorname{span}_k\{[a,b]_{\mathfrak g} : a \in A,\ b \in B\}. \end{align*} Because $I \trianglelefteq \mathfrak g$, the quotient [vector space](/page/Vector%20Space) $\mathfrak g/I$ is a Lie algebra with bracket \begin{align*} [\cdot,\cdot]_{\mathfrak g/I}: (\mathfrak g/I)\times(\mathfrak g/I) &\to \mathfrak g/I,\\ (x+I,y+I) &\mapsto [x,y]_{\mathfrak g}+I. \end{align*} This is well-defined because if $x'-x \in I$ and $y'-y \in I$, then \begin{align*} [x',y']_{\mathfrak g}-[x,y]_{\mathfrak g} &= [x'-x,y]_{\mathfrak g}+[x,y'-y]_{\mathfrak g}+[x'-x,y'-y]_{\mathfrak g} \in I, \end{align*} using $[I,\mathfrak g]_{\mathfrak g}\subseteq I$, $[\mathfrak g,I]_{\mathfrak g}\subseteq I$, and $[I,I]_{\mathfrak g}\subseteq I$. [/step]

[step:Prove the lower central series formula by induction]We prove that \begin{align*} \gamma_i(\mathfrak g/I)=\frac{\gamma_i(\mathfrak g)+I}{I} \end{align*} for every $i \in \mathbb N$. For $i=1$, \begin{align*} \gamma_1(\mathfrak g/I)=\mathfrak g/I=\frac{\mathfrak g+I}{I}=\frac{\gamma_1(\mathfrak g)+I}{I}. \end{align*} Assume the formula holds for some $i \in \mathbb N$. Then \begin{align*} \gamma_{i+1}(\mathfrak g/I) &= [\mathfrak g/I,\gamma_i(\mathfrak g/I)]_{\mathfrak g/I}\\ &= \left[\mathfrak g/I,\frac{\gamma_i(\mathfrak g)+I}{I}\right]_{\mathfrak g/I}. \end{align*} By the quotient bracket, the right-hand side is \begin{align*} \frac{[\mathfrak g,\gamma_i(\mathfrak g)]+I}{I} = \frac{\gamma_{i+1}(\mathfrak g)+I}{I}. \end{align*} Thus the formula holds for $i+1$, and induction gives the result for all $i \in \mathbb N$.[/step]

[guided]We prove the identity by induction because the lower central series is itself defined recursively. The claim for each $i \in \mathbb N$ is \begin{align*} \gamma_i(\mathfrak g/I)=\frac{\gamma_i(\mathfrak g)+I}{I}. \end{align*} For the base case $i=1$, the definition of the lower central series gives $\gamma_1(\mathfrak g/I)=\mathfrak g/I$ and $\gamma_1(\mathfrak g)=\mathfrak g$. Therefore \begin{align*} \gamma_1(\mathfrak g/I)=\mathfrak g/I=\frac{\mathfrak g+I}{I}=\frac{\gamma_1(\mathfrak g)+I}{I}. \end{align*} Now fix $i \in \mathbb N$ and assume \begin{align*} \gamma_i(\mathfrak g/I)=\frac{\gamma_i(\mathfrak g)+I}{I}. \end{align*} Using the recursive definition of the lower central series in the quotient Lie algebra, \begin{align*} \gamma_{i+1}(\mathfrak g/I) &= [\mathfrak g/I,\gamma_i(\mathfrak g/I)]_{\mathfrak g/I}\\ &= \left[\mathfrak g/I,\frac{\gamma_i(\mathfrak g)+I}{I}\right]_{\mathfrak g/I}. \end{align*} We now compute this quotient bracket. Every element of $\mathfrak g/I$ has the form $x+I$ with $x \in \mathfrak g$, and every element of $(\gamma_i(\mathfrak g)+I)/I$ has the form $y+I$ with $y \in \gamma_i(\mathfrak g)$. Their bracket in the quotient is \begin{align*} [x+I,y+I]_{\mathfrak g/I}=[x,y]_{\mathfrak g}+I. \end{align*} Taking the $k$-linear span over all such $x$ and $y$ gives \begin{align*} \left[\mathfrak g/I,\frac{\gamma_i(\mathfrak g)+I}{I}\right]_{\mathfrak g/I} = \frac{[\mathfrak g,\gamma_i(\mathfrak g)]+I}{I}. \end{align*} By the definition of the lower central series in $\mathfrak g$, \begin{align*} [\mathfrak g,\gamma_i(\mathfrak g)] = \gamma_{i+1}(\mathfrak g). \end{align*} Hence \begin{align*} \gamma_{i+1}(\mathfrak g/I) = \frac{\gamma_{i+1}(\mathfrak g)+I}{I}. \end{align*} This proves the induction step, so the identity holds for every $i \in \mathbb N$.[/guided]

[step:Deduce nilpotence of the quotient] Assume $\mathfrak g$ is nilpotent. Then there exists $c \in \mathbb N$ such that \begin{align*} \gamma_{c+1}(\mathfrak g)=0. \end{align*} Using the formula already proved with $i=c+1$, \begin{align*} \gamma_{c+1}(\mathfrak g/I) = \frac{\gamma_{c+1}(\mathfrak g)+I}{I} = \frac{0+I}{I} = I/I = 0. \end{align*} Therefore the lower central series of $\mathfrak g/I$ reaches $0$, so $\mathfrak g/I$ is nilpotent. [/step]