[proofplan] We identify each lower central term $\gamma_m(\mathfrak g)$ with the span of right-nested iterated brackets of length $m$. This is proved by induction directly from the recursive definition $\gamma_{m+1}(\mathfrak g)=[\mathfrak g,\gamma_m(\mathfrak g)]$ and bilinearity of the Lie bracket. Once this spanning description is established for $m=c+1$, the vanishing of $\gamma_{c+1}(\mathfrak g)$ is exactly the statement that every such iterated bracket of length $c+1$ is zero. [/proofplan]

[step:Describe lower central terms by right-nested brackets]For each $m \in \mathbb N$, define $R_m \subseteq \mathfrak g$ by \begin{align*} R_m := \operatorname{span}_k \left\{ [x_1,[x_2,[\cdots,[x_{m-1},x_m]\cdots]]] : x_1,\dots,x_m \in \mathfrak g \right\}, \end{align*} with the convention that $R_1=\operatorname{span}_k\{x_1:x_1\in\mathfrak g\}=\mathfrak g$. We prove by induction on $m \in \mathbb N$ that \begin{align*} \gamma_m(\mathfrak g)=R_m. \end{align*} For $m=1$, both spaces are equal to $\mathfrak g$. Assume that $\gamma_m(\mathfrak g)=R_m$ for some $m \in \mathbb N$. By definition of the lower central series, \begin{align*} \gamma_{m+1}(\mathfrak g) &= [\mathfrak g,\gamma_m(\mathfrak g)] \\ &= [\mathfrak g,R_m]. \end{align*} Since the Lie bracket $[\cdot,\cdot]:\mathfrak g\times\mathfrak g\to\mathfrak g$ is $k$-bilinear, the subspace $[\mathfrak g,R_m]$ is the $k$-linear span of all elements \begin{align*} [x_0,[x_1,[x_2,[\cdots,[x_{m-1},x_m]\cdots]]]] \end{align*} with $x_0,x_1,\dots,x_m \in \mathfrak g$. Renaming $x_0,x_1,\dots,x_m$ as $y_1,y_2,\dots,y_{m+1}$ gives precisely $R_{m+1}$. Hence $\gamma_{m+1}(\mathfrak g)=R_{m+1}$, completing the induction.[/step]

[guided]The lower central series is defined recursively: each new term is obtained by bracketing one arbitrary element of $\mathfrak g$ with one element from the previous term. We make that recursive structure explicit. For each $m \in \mathbb N$, define $R_m$ to be the $k$-linear span of all right-nested brackets of length $m$: \begin{align*} R_m := \operatorname{span}_k \left\{ [x_1,[x_2,[\cdots,[x_{m-1},x_m]\cdots]]] : x_1,\dots,x_m \in \mathfrak g \right\}. \end{align*} When $m=1$, there is no bracket; the convention is \begin{align*} R_1=\operatorname{span}_k\{x_1:x_1\in\mathfrak g\}=\mathfrak g. \end{align*} We prove $\gamma_m(\mathfrak g)=R_m$ by induction on $m$. The base case is immediate from the definitions: \begin{align*} \gamma_1(\mathfrak g)=\mathfrak g=R_1. \end{align*} Now assume that $\gamma_m(\mathfrak g)=R_m$ for some fixed $m \in \mathbb N$. The recursive definition of the lower central series gives \begin{align*} \gamma_{m+1}(\mathfrak g) = [\mathfrak g,\gamma_m(\mathfrak g)]. \end{align*} Using the induction hypothesis, this becomes \begin{align*} \gamma_{m+1}(\mathfrak g) = [\mathfrak g,R_m]. \end{align*} The notation $[\mathfrak g,R_m]$ means the $k$-linear span of all brackets $[x,y]$ with $x\in\mathfrak g$ and $y\in R_m$. Since $R_m$ is itself spanned by the right-nested brackets \begin{align*} [x_1,[x_2,[\cdots,[x_{m-1},x_m]\cdots]]], \end{align*} and since the Lie bracket is $k$-bilinear, bracketing an arbitrary $x_0\in\mathfrak g$ with a spanning element of $R_m$ produces exactly the brackets \begin{align*} [x_0,[x_1,[x_2,[\cdots,[x_{m-1},x_m]\cdots]]]] \end{align*} with $x_0,x_1,\dots,x_m\in\mathfrak g$. Their $k$-linear span is $R_{m+1}$ after renaming the variables as $y_1,\dots,y_{m+1}$. Therefore \begin{align*} \gamma_{m+1}(\mathfrak g)=R_{m+1}. \end{align*} This proves the spanning description for every $m \in \mathbb N$.[/guided]

[step:Translate vanishing of $\gamma_{c+1}(\mathfrak g)$ into vanishing of all length $c+1$ brackets] By the previous step with $m=c+1$, \begin{align*} \gamma_{c+1}(\mathfrak g) = \operatorname{span}_k \left\{ [x_1,[x_2,[\cdots,[x_c,x_{c+1}]\cdots]]] : x_1,\dots,x_{c+1}\in\mathfrak g \right\}. \end{align*} If $\mathfrak g$ is nilpotent of class at most $c$, then $\gamma_{c+1}(\mathfrak g)=0$. Every generator of the displayed spanning set lies in $\gamma_{c+1}(\mathfrak g)$, so every bracket \begin{align*} [x_1,[x_2,[\cdots,[x_c,x_{c+1}]\cdots]]] \end{align*} is zero. Conversely, assume that \begin{align*} [x_1,[x_2,[\cdots,[x_c,x_{c+1}]\cdots]]]=0 \end{align*} for all $x_1,\dots,x_{c+1}\in\mathfrak g$. Then every generator of the displayed spanning set for $\gamma_{c+1}(\mathfrak g)$ is zero, so its $k$-linear span is the zero subspace: \begin{align*} \gamma_{c+1}(\mathfrak g)=0. \end{align*} Thus $\mathfrak g$ is nilpotent of class at most $c$. This proves both implications. [/step]