[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.
[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]
[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]
