[proofplan] We prove both directions by comparing an arbitrary central series with the lower central series. If the lower central series reaches $0$, then reversing its nonzero part gives a chain whose commutator condition is exactly the defining recurrence of the lower central series. Conversely, if a central series is given, an induction shows that each lower central term $\gamma_i(\mathfrak g)$ is forced into the corresponding earlier term $\mathfrak g_{c+1-i}$ of the central series; at the last index this gives $\gamma_{c+1}(\mathfrak g)=0$. [/proofplan]

[step:Construct a central series from a vanishing lower central series]Assume that $\mathfrak g$ is nilpotent. Then there exists an integer $c \geq 0$ such that $\gamma_{c+1}(\mathfrak g)=0$. If $c=0$, then $\gamma_1(\mathfrak g)=\mathfrak g=0$, and the chain $0=\mathfrak g_0=\mathfrak g$ is a central series. Now assume $c \geq 1$. For each integer $0 \leq i \leq c$, define the linear subspace \begin{align*} \mathfrak g_i := \gamma_{c+1-i}(\mathfrak g). \end{align*} Then $\mathfrak g_0=\gamma_{c+1}(\mathfrak g)=0$ and $\mathfrak g_c=\gamma_1(\mathfrak g)=\mathfrak g$. We verify that this is an increasing chain of ideals. First, each $\gamma_j(\mathfrak g)$ is an ideal of $\mathfrak g$. Indeed, $\gamma_1(\mathfrak g)=\mathfrak g$ is an ideal. If $\gamma_j(\mathfrak g)$ is an ideal, then for $x,y \in \mathfrak g$ and $z \in \gamma_j(\mathfrak g)$, the Jacobi identity gives \begin{align*} [x,[y,z]] = [[x,y],z] + [y,[x,z]]. \end{align*} Since $[x,y]\in \mathfrak g$, $z\in \gamma_j(\mathfrak g)$, and $[x,z]\in \gamma_j(\mathfrak g)$, both terms on the right belong to $[\mathfrak g,\gamma_j(\mathfrak g)]=\gamma_{j+1}(\mathfrak g)$. Hence $[\mathfrak g,\gamma_{j+1}(\mathfrak g)]\subseteq \gamma_{j+1}(\mathfrak g)$, so $\gamma_{j+1}(\mathfrak g)$ is an ideal. Also, because each $\gamma_j(\mathfrak g)$ is an ideal, \begin{align*} \gamma_{j+1}(\mathfrak g)=[\mathfrak g,\gamma_j(\mathfrak g)]\subseteq \gamma_j(\mathfrak g). \end{align*} Thus \begin{align*} 0=\mathfrak g_0 \subseteq \mathfrak g_1 \subseteq \cdots \subseteq \mathfrak g_c=\mathfrak g. \end{align*} Finally, for each integer $1 \leq i \leq c$, \begin{align*} [\mathfrak g,\mathfrak g_i] = [\mathfrak g,\gamma_{c+1-i}(\mathfrak g)] = \gamma_{c+2-i}(\mathfrak g) = \mathfrak g_{i-1}. \end{align*} Therefore the chain is a central series.[/step]

[guided]Assume that $\mathfrak g$ is nilpotent. By definition, this means that the lower central series eventually reaches $0$: there is an integer $c \geq 0$ such that $\gamma_{c+1}(\mathfrak g)=0$. If $c=0$, then $\gamma_1(\mathfrak g)=\mathfrak g=0$. In that case the one-term chain $0=\mathfrak g_0=\mathfrak g$ satisfies the required condition, since there are no indices $i$ with $1 \leq i \leq 0$. Now suppose $c \geq 1$. The central series should climb from $0$ up to $\mathfrak g$, while the lower central series descends from $\mathfrak g$ down to $0$. Therefore we reverse the lower central series. For each integer $0 \leq i \leq c$, define \begin{align*} \mathfrak g_i := \gamma_{c+1-i}(\mathfrak g). \end{align*} This gives the correct endpoints: \begin{align*} \mathfrak g_0=\gamma_{c+1}(\mathfrak g)=0, \qquad \mathfrak g_c=\gamma_1(\mathfrak g)=\mathfrak g. \end{align*} We must still verify that these subspaces are ideals and that they form an increasing chain. First we prove that every lower central term is an ideal of $\mathfrak g$. The first term $\gamma_1(\mathfrak g)=\mathfrak g$ is an ideal. Suppose $\gamma_j(\mathfrak g)$ is an ideal. We must show that $\gamma_{j+1}(\mathfrak g)=[\mathfrak g,\gamma_j(\mathfrak g)]$ is stable under bracketing with arbitrary elements of $\mathfrak g$. It is enough to check generators of $[\mathfrak g,\gamma_j(\mathfrak g)]$. Let $x,y \in \mathfrak g$ and let $z \in \gamma_j(\mathfrak g)$. The Jacobi identity gives \begin{align*} [x,[y,z]] = [[x,y],z] + [y,[x,z]]. \end{align*} Because $[x,y]\in \mathfrak g$ and $z\in \gamma_j(\mathfrak g)$, the term $[[x,y],z]$ lies in $[\mathfrak g,\gamma_j(\mathfrak g)]$. Because $\gamma_j(\mathfrak g)$ is an ideal, $[x,z]\in \gamma_j(\mathfrak g)$, so $[y,[x,z]]$ also lies in $[\mathfrak g,\gamma_j(\mathfrak g)]$. Hence $[x,[y,z]]\in \gamma_{j+1}(\mathfrak g)$, proving that $\gamma_{j+1}(\mathfrak g)$ is an ideal. Since each $\gamma_j(\mathfrak g)$ is an ideal, it is stable under brackets with elements of $\mathfrak g$. Therefore \begin{align*} \gamma_{j+1}(\mathfrak g)=[\mathfrak g,\gamma_j(\mathfrak g)]\subseteq \gamma_j(\mathfrak g). \end{align*} Thus the lower central series is decreasing. Reversing a decreasing chain gives an increasing chain, so \begin{align*} 0=\mathfrak g_0 \subseteq \mathfrak g_1 \subseteq \cdots \subseteq \mathfrak g_c=\mathfrak g. \end{align*} It remains to check the centrality condition. For each integer $1 \leq i \leq c$, the definition of $\mathfrak g_i$ and the recurrence defining the lower central series give \begin{align*} [\mathfrak g,\mathfrak g_i] = [\mathfrak g,\gamma_{c+1-i}(\mathfrak g)] = \gamma_{c+2-i}(\mathfrak g) = \mathfrak g_{i-1}. \end{align*} Thus $[\mathfrak g,\mathfrak g_i]\subseteq \mathfrak g_{i-1}$ for every $1 \leq i \leq c$, so the reversed lower central series is a central series.[/guided]

[step:Force the lower central series into any central series]Conversely, suppose that there exist an integer $c \geq 0$ and ideals \begin{align*} 0=\mathfrak g_0 \subseteq \mathfrak g_1 \subseteq \cdots \subseteq \mathfrak g_c=\mathfrak g \end{align*} such that $[\mathfrak g,\mathfrak g_i]\subseteq \mathfrak g_{i-1}$ for every integer $1 \leq i \leq c$. We prove by induction on $i$ that \begin{align*} \gamma_i(\mathfrak g)\subseteq \mathfrak g_{c+1-i} \end{align*} for every integer $1 \leq i \leq c+1$. For $i=1$, \begin{align*} \gamma_1(\mathfrak g)=\mathfrak g=\mathfrak g_c, \end{align*} so the assertion holds. Assume that $1 \leq i \leq c$ and that $\gamma_i(\mathfrak g)\subseteq \mathfrak g_{c+1-i}$. Then $1 \leq c+1-i \leq c$, so the centrality condition applies to $\mathfrak g_{c+1-i}$. Using the definition of the lower central series and the induction hypothesis, \begin{align*} \gamma_{i+1}(\mathfrak g) = [\mathfrak g,\gamma_i(\mathfrak g)] \subseteq [\mathfrak g,\mathfrak g_{c+1-i}] \subseteq \mathfrak g_{c-i}. \end{align*} Since $c-i=c+1-(i+1)$, this is exactly \begin{align*} \gamma_{i+1}(\mathfrak g)\subseteq \mathfrak g_{c+1-(i+1)}. \end{align*} The induction is complete. Taking $i=c+1$ gives \begin{align*} \gamma_{c+1}(\mathfrak g)\subseteq \mathfrak g_0=0. \end{align*} Hence $\gamma_{c+1}(\mathfrak g)=0$, so $\mathfrak g$ is nilpotent.[/step]

[guided]Now suppose a central series is given. Thus there are an integer $c \geq 0$ and ideals \begin{align*} 0=\mathfrak g_0 \subseteq \mathfrak g_1 \subseteq \cdots \subseteq \mathfrak g_c=\mathfrak g \end{align*} such that \begin{align*} [\mathfrak g,\mathfrak g_i]\subseteq \mathfrak g_{i-1} \end{align*} for every integer $1 \leq i \leq c$. The goal is to prove that the lower central series reaches $0$. The central series says that bracketing with $\mathfrak g$ moves one step downward in the chain. Since the lower central series is built by repeatedly bracketing with $\mathfrak g$, it should move downward one step at a time through the central series. We make this precise by induction. We prove that \begin{align*} \gamma_i(\mathfrak g)\subseteq \mathfrak g_{c+1-i} \end{align*} for every integer $1 \leq i \leq c+1$. For $i=1$, the assertion is \begin{align*} \gamma_1(\mathfrak g)\subseteq \mathfrak g_c. \end{align*} But $\gamma_1(\mathfrak g)=\mathfrak g$ by definition and $\mathfrak g_c=\mathfrak g$ by the endpoint condition, so \begin{align*} \gamma_1(\mathfrak g)=\mathfrak g=\mathfrak g_c. \end{align*} Assume now that $1 \leq i \leq c$ and that \begin{align*} \gamma_i(\mathfrak g)\subseteq \mathfrak g_{c+1-i}. \end{align*} We need to prove \begin{align*} \gamma_{i+1}(\mathfrak g)\subseteq \mathfrak g_{c+1-(i+1)}. \end{align*} Because $1 \leq i \leq c$, the index $c+1-i$ satisfies $1 \leq c+1-i \leq c$, so the centrality condition applies to $\mathfrak g_{c+1-i}$. Using the definition of the lower central series, then the induction hypothesis, and then the centrality condition, we obtain \begin{align*} \gamma_{i+1}(\mathfrak g) = [\mathfrak g,\gamma_i(\mathfrak g)] \subseteq [\mathfrak g,\mathfrak g_{c+1-i}] \subseteq \mathfrak g_{c-i}. \end{align*} Since \begin{align*} c-i=c+1-(i+1), \end{align*} this is exactly the desired inclusion \begin{align*} \gamma_{i+1}(\mathfrak g)\subseteq \mathfrak g_{c+1-(i+1)}. \end{align*} By induction, the inclusion holds for every integer $1 \leq i \leq c+1$. Substituting $i=c+1$ gives \begin{align*} \gamma_{c+1}(\mathfrak g)\subseteq \mathfrak g_0=0. \end{align*} Therefore $\gamma_{c+1}(\mathfrak g)=0$. This is precisely the definition that $\mathfrak g$ is nilpotent.[/guided]

[step:Conclude the equivalence] The first step proves that nilpotence implies the existence of a central series. The second step proves that the existence of a central series implies nilpotence. Therefore a Lie algebra $\mathfrak g$ is nilpotent if and only if it admits a central series. [/step]