[proofplan] We compare the lower central series of $\mathfrak g$ with the lower central series of the quotient $\mathfrak g/I$ using the quotient map. The key point is that the image of $\gamma_m(\mathfrak g)$ in the quotient is exactly $\gamma_m(\mathfrak g/I)$ for every $m \geq 1$. Once nilpotency of the quotient forces some $\gamma_{c+1}(\mathfrak g)$ to lie inside $I$, centrality of $I$ makes the next lower central term vanish. [/proofplan] [step:Relate the lower central series of $\mathfrak g$ to that of the quotient] Let \begin{align*} \pi:\mathfrak g &\to \mathfrak g/I \\ x &\mapsto x+I \end{align*} be the quotient Lie algebra homomorphism. We prove that for every integer $m \geq 1$, \begin{align*} \gamma_m(\mathfrak g/I)=\pi(\gamma_m(\mathfrak g))=\frac{\gamma_m(\mathfrak g)+I}{I}. \end{align*} For $m=1$, this is \begin{align*} \gamma_1(\mathfrak g/I)=\mathfrak g/I=\pi(\mathfrak g)=\pi(\gamma_1(\mathfrak g)). \end{align*} Assume the identity holds for some integer $m \geq 1$. Since $\pi$ is a Lie algebra homomorphism, it preserves brackets: \begin{align*} \pi([x,y])=[\pi(x),\pi(y)] \end{align*} for all $x,y \in \mathfrak g$. Therefore \begin{align*} \gamma_{m+1}(\mathfrak g/I) &=[\mathfrak g/I,\gamma_m(\mathfrak g/I)] \\ &=[\pi(\mathfrak g),\pi(\gamma_m(\mathfrak g))] \\ &=\pi([\mathfrak g,\gamma_m(\mathfrak g)]) \\ &=\pi(\gamma_{m+1}(\mathfrak g)). \end{align*} This completes the induction. The equality $\pi(\gamma_m(\mathfrak g))=(\gamma_m(\mathfrak g)+I)/I$ is the standard description of the image of a subspace under the quotient map. [guided] The quotient map \begin{align*} \pi:\mathfrak g &\to \mathfrak g/I \\ x &\mapsto x+I \end{align*} is a surjective Lie algebra homomorphism because $I$ is an ideal of $\mathfrak g$. We want to know how the lower central series behaves under this map. The claim is that the $m$-th lower central term of the quotient is exactly the image of the $m$-th lower central term of $\mathfrak g$: \begin{align*} \gamma_m(\mathfrak g/I)=\pi(\gamma_m(\mathfrak g))=\frac{\gamma_m(\mathfrak g)+I}{I}. \end{align*} We prove this by induction on $m$. For $m=1$, the lower central series starts with the entire Lie algebra, so \begin{align*} \gamma_1(\mathfrak g/I)=\mathfrak g/I=\pi(\mathfrak g)=\pi(\gamma_1(\mathfrak g)). \end{align*} Now assume the formula holds for some integer $m \geq 1$. The next lower central term is obtained by bracketing with the whole Lie algebra. Since $\pi$ is a Lie algebra homomorphism, it satisfies \begin{align*} \pi([x,y])=[\pi(x),\pi(y)] \end{align*} for all $x,y \in \mathfrak g$. Using the induction hypothesis, we compute: \begin{align*} \gamma_{m+1}(\mathfrak g/I) &=[\mathfrak g/I,\gamma_m(\mathfrak g/I)] \\ &=[\pi(\mathfrak g),\pi(\gamma_m(\mathfrak g))] \\ &=\pi([\mathfrak g,\gamma_m(\mathfrak g)]) \\ &=\pi(\gamma_{m+1}(\mathfrak g)). \end{align*} The third equality holds because the bracket of the two image subspaces is spanned by brackets of the form $[\pi(x),\pi(y)]$, and each such bracket equals $\pi([x,y])$. Finally, for any subspace $A \subseteq \mathfrak g$, its image under the quotient map is $(A+I)/I$, so \begin{align*} \pi(\gamma_m(\mathfrak g))=\frac{\gamma_m(\mathfrak g)+I}{I}. \end{align*} Thus the quotient lower central series is exactly the image of the original lower central series. [/guided] [/step] [step:Use nilpotency of the quotient to force a lower central term into the central ideal] Assume $\gamma_{c+1}(\mathfrak g/I)=0$ for some integer $c \geq 0$. By the quotient formula from the previous step, \begin{align*} 0=\gamma_{c+1}(\mathfrak g/I)=\frac{\gamma_{c+1}(\mathfrak g)+I}{I}. \end{align*} Hence $\gamma_{c+1}(\mathfrak g)+I=I$, and therefore \begin{align*} \gamma_{c+1}(\mathfrak g)\subseteq I. \end{align*} [/step] [step:Bracket once more and use centrality to terminate the lower central series] Since $I \subseteq Z(\mathfrak g)$, every element of $I$ brackets to zero with every element of $\mathfrak g$. Therefore \begin{align*} [\mathfrak g,I]=0. \end{align*} Using $\gamma_{c+1}(\mathfrak g)\subseteq I$, we obtain \begin{align*} \gamma_{c+2}(\mathfrak g) &=[\mathfrak g,\gamma_{c+1}(\mathfrak g)] \\ &\subseteq [\mathfrak g,I] \\ &=0. \end{align*} Thus the lower central series of $\mathfrak g$ terminates. Hence $\mathfrak g$ is nilpotent, with nilpotency class at most $c+1$. [/step]