[proofplan] We prove by induction on $i$ that $\gamma_i(\mathfrak g)$ is an ideal of $\mathfrak g$. The base case is $\gamma_1(\mathfrak g)=\mathfrak g$. For the inductive step, we use the Jacobi identity to show that bracketing an element of $\mathfrak g$ with a generator $[y,z]$ of $[\mathfrak g,\gamma_i(\mathfrak g)]$ again lands in $[\mathfrak g,\gamma_i(\mathfrak g)]$. Once the ideal property is known, the containment $\gamma_{i+1}(\mathfrak g)\subseteq \gamma_i(\mathfrak g)$ follows directly from the definition of ideal. [/proofplan]

[step:Initialize the induction at $\gamma_1(\mathfrak g)=\mathfrak g$] The subspace $\gamma_1(\mathfrak g)$ is equal to $\mathfrak g$ by definition. Since $[\mathfrak g,\mathfrak g]\subseteq \mathfrak g$, the subspace $\gamma_1(\mathfrak g)$ is an ideal of $\mathfrak g$. [/step]

[step:Use the Jacobi identity to prove the next lower central term is an ideal]Assume that $i\geq 1$ and that $\gamma_i(\mathfrak g)$ is an ideal of $\mathfrak g$. We prove that $\gamma_{i+1}(\mathfrak g)$ is an ideal. By definition, \begin{align*} \gamma_{i+1}(\mathfrak g)=[\mathfrak g,\gamma_i(\mathfrak g)]. \end{align*} Thus $\gamma_{i+1}(\mathfrak g)$ is the $k$-linear span of brackets $[y,z]$ with $y\in \mathfrak g$ and $z\in \gamma_i(\mathfrak g)$. To prove that it is an ideal, it is enough by bilinearity of the Lie bracket to show that \begin{align*} [x,[y,z]]\in \gamma_{i+1}(\mathfrak g) \end{align*} for all $x,y\in \mathfrak g$ and all $z\in \gamma_i(\mathfrak g)$. Let $x,y\in \mathfrak g$ and $z\in \gamma_i(\mathfrak g)$. The Jacobi identity gives \begin{align*} [x,[y,z]] = [[x,y],z] + [y,[x,z]]. \end{align*} Since $[x,y]\in \mathfrak g$ and $z\in \gamma_i(\mathfrak g)$, the term $[[x,y],z]$ lies in $[\mathfrak g,\gamma_i(\mathfrak g)] = \gamma_{i+1}(\mathfrak g)$. Since $\gamma_i(\mathfrak g)$ is an ideal and $x\in \mathfrak g$, we also have $[x,z]\in \gamma_i(\mathfrak g)$; hence $[y,[x,z]]\in [\mathfrak g,\gamma_i(\mathfrak g)] = \gamma_{i+1}(\mathfrak g)$. Therefore $[x,[y,z]]\in \gamma_{i+1}(\mathfrak g)$. Because every element of $\gamma_{i+1}(\mathfrak g)$ is a finite $k$-linear combination of such generators $[y,z]$, bilinearity implies \begin{align*} [\mathfrak g,\gamma_{i+1}(\mathfrak g)]\subseteq \gamma_{i+1}(\mathfrak g). \end{align*} Thus $\gamma_{i+1}(\mathfrak g)$ is an ideal of $\mathfrak g$.[/step]

[guided]Assume that $i\geq 1$ and that $\gamma_i(\mathfrak g)$ is already known to be an ideal. We need to prove that the next term \begin{align*} \gamma_{i+1}(\mathfrak g)=[\mathfrak g,\gamma_i(\mathfrak g)] \end{align*} is also stable under bracketing with arbitrary elements of $\mathfrak g$. The elements of $\gamma_{i+1}(\mathfrak g)$ are finite $k$-linear combinations of brackets $[y,z]$, where $y\in \mathfrak g$ and $z\in \gamma_i(\mathfrak g)$. Since the Lie bracket is bilinear, it is enough to check stability on these spanning generators. Thus fix $x,y\in \mathfrak g$ and $z\in \gamma_i(\mathfrak g)$, and consider $[x,[y,z]]$. The Jacobi identity rewrites this bracket as \begin{align*} [x,[y,z]] = [[x,y],z] + [y,[x,z]]. \end{align*} Now each term lies in $\gamma_{i+1}(\mathfrak g)$. For the first term, $[x,y]\in \mathfrak g$ and $z\in \gamma_i(\mathfrak g)$, so \begin{align*} [[x,y],z]\in [\mathfrak g,\gamma_i(\mathfrak g)] = \gamma_{i+1}(\mathfrak g). \end{align*} For the second term, the induction hypothesis is used: because $\gamma_i(\mathfrak g)$ is an ideal, $[x,z]\in \gamma_i(\mathfrak g)$. Since $y\in \mathfrak g$, this gives \begin{align*} [y,[x,z]]\in [\mathfrak g,\gamma_i(\mathfrak g)] = \gamma_{i+1}(\mathfrak g). \end{align*} Therefore $[x,[y,z]]$ is a sum of two elements of $\gamma_{i+1}(\mathfrak g)$, and hence belongs to $\gamma_{i+1}(\mathfrak g)$. Finally, because arbitrary elements of $\gamma_{i+1}(\mathfrak g)$ are finite $k$-linear combinations of the generators $[y,z]$, bilinearity of the bracket gives \begin{align*} [\mathfrak g,\gamma_{i+1}(\mathfrak g)]\subseteq \gamma_{i+1}(\mathfrak g). \end{align*} This is exactly the statement that $\gamma_{i+1}(\mathfrak g)$ is an ideal of $\mathfrak g$.[/guided]

[step:Conclude ideality for all lower central terms by induction] The base case proves that $\gamma_1(\mathfrak g)$ is an ideal. The inductive step proves that if $\gamma_i(\mathfrak g)$ is an ideal, then $\gamma_{i+1}(\mathfrak g)$ is an ideal. Therefore, by induction on $i$, the subspace $\gamma_i(\mathfrak g)$ is an ideal of $\mathfrak g$ for every integer $i\geq 1$. [/step]

[step:Deduce that the lower central series is descending] Let $i\geq 1$. Since $\gamma_i(\mathfrak g)$ is an ideal of $\mathfrak g$, it satisfies \begin{align*} [\mathfrak g,\gamma_i(\mathfrak g)]\subseteq \gamma_i(\mathfrak g). \end{align*} Using the definition of the next lower central term, \begin{align*} \gamma_{i+1}(\mathfrak g)=[\mathfrak g,\gamma_i(\mathfrak g)]\subseteq \gamma_i(\mathfrak g). \end{align*} This proves the stated containment for every integer $i\geq 1$ and completes the proof. [/step]