[proofplan] We compare the derived series of $\mathfrak g$ with its lower central series. The key estimate is that brackets of lower central terms satisfy $[\gamma_a(\mathfrak g), \gamma_b(\mathfrak g)] \subseteq \gamma_{a+b}(\mathfrak g)$, proved by induction from the Jacobi identity. This estimate implies $\mathfrak g^{(r)} \subseteq \gamma_{2^r}(\mathfrak g)$ for every $r \geq 0$. If the lower central series vanishes, then a sufficiently far term of the derived series also vanishes, which is exactly solvability. [/proofplan]

[step:Define the two descending series and the bracket notation] For two $k$-linear subspaces $A, B \subseteq \mathfrak g$, define \begin{align*} [A,B] := \operatorname{span}_k\{[x,y] : x \in A,\ y \in B\}. \end{align*} Define the lower central series $\gamma_m(\mathfrak g)$ by \begin{align*} \gamma_1(\mathfrak g) &:= \mathfrak g,\\ \gamma_{m+1}(\mathfrak g) &:= [\mathfrak g,\gamma_m(\mathfrak g)] \qquad \text{for } m \geq 1. \end{align*} Define the derived series $\mathfrak g^{(r)}$ by \begin{align*} \mathfrak g^{(0)} &:= \mathfrak g,\\ \mathfrak g^{(r+1)} &:= [\mathfrak g^{(r)},\mathfrak g^{(r)}] \qquad \text{for } r \geq 0. \end{align*} The Lie algebra $\mathfrak g$ is nilpotent if there exists an integer $N \geq 1$ such that $\gamma_N(\mathfrak g)=0$, and it is solvable if there exists an integer $R \geq 0$ such that $\mathfrak g^{(R)}=0$. [/step]

[step:Prove that brackets of lower central terms move deeper in the lower central series][claim:Lower central bracket estimate] For all integers $a,b \geq 1$, \begin{align*} [\gamma_a(\mathfrak g),\gamma_b(\mathfrak g)] \subseteq \gamma_{a+b}(\mathfrak g). \end{align*} [/claim] [proof] We prove the claim by induction on the first index $a$, with the assertion for a fixed $a$ required to hold for every integer $b \geq 1$. For $a=1$ and every $b \geq 1$, the definition of the lower central series gives \begin{align*} [\gamma_1(\mathfrak g),\gamma_b(\mathfrak g)] = [\mathfrak g,\gamma_b(\mathfrak g)] = \gamma_{b+1}(\mathfrak g) = \gamma_{1+b}(\mathfrak g). \end{align*} Thus the assertion holds for $a=1$. Assume now that $a \geq 2$ and that, for every integer $c \geq 1$, the containment \begin{align*} [\gamma_{a-1}(\mathfrak g),\gamma_c(\mathfrak g)] \subseteq \gamma_{a-1+c}(\mathfrak g) \end{align*} has already been proved. Let $b \geq 1$. Since \begin{align*} \gamma_a(\mathfrak g)=[\mathfrak g,\gamma_{a-1}(\mathfrak g)], \end{align*} it is enough by bilinearity of the Lie bracket to prove that $[[x,u],v] \in \gamma_{a+b}(\mathfrak g)$ for arbitrary elements \begin{align*} x \in \mathfrak g, \qquad u \in \gamma_{a-1}(\mathfrak g), \qquad v \in \gamma_b(\mathfrak g). \end{align*} The Jacobi identity gives \begin{align*} [[x,u],v]=[x,[u,v]]-[u,[x,v]]. \end{align*} By the induction hypothesis with $c=b$, \begin{align*} [u,v] \in [\gamma_{a-1}(\mathfrak g),\gamma_b(\mathfrak g)] \subseteq \gamma_{a+b-1}(\mathfrak g), \end{align*} and therefore, by the defining recursion for the lower central series, \begin{align*} [x,[u,v]] \in [\mathfrak g,\gamma_{a+b-1}(\mathfrak g)] = \gamma_{a+b}(\mathfrak g). \end{align*} Also, because $x \in \mathfrak g=\gamma_1(\mathfrak g)$ and $v \in \gamma_b(\mathfrak g)$, \begin{align*} [x,v] \in [\gamma_1(\mathfrak g),\gamma_b(\mathfrak g)] = \gamma_{b+1}(\mathfrak g). \end{align*} Applying the induction hypothesis again, now with $c=b+1$, gives \begin{align*} [u,[x,v]] \in [\gamma_{a-1}(\mathfrak g),\gamma_{b+1}(\mathfrak g)] \subseteq \gamma_{a+b}(\mathfrak g). \end{align*} Since $\gamma_{a+b}(\mathfrak g)$ is a $k$-linear subspace, the difference \begin{align*} [x,[u,v]]-[u,[x,v]] \end{align*} also lies in $\gamma_{a+b}(\mathfrak g)$. Hence $[[x,u],v] \in \gamma_{a+b}(\mathfrak g)$, and the induction step is complete. Therefore the claim holds for all integers $a,b \geq 1$. [/proof][/step]

[guided]The point of this estimate is that the lower central index behaves additively under brackets. The induction must be arranged carefully: induction on the total index $a+b$ would be circular, because the Jacobi identity naturally produces the pair $(a-1,b+1)$, which has the same total index. Instead, we induct on the first index $a$, and at each fixed value of $a$ we prove the assertion simultaneously for every $b \geq 1$. For $a=1$ and any integer $b \geq 1$, the statement is exactly the recursive definition of the lower central series. Since $\gamma_1(\mathfrak g)=\mathfrak g$, we have \begin{align*} [\gamma_1(\mathfrak g),\gamma_b(\mathfrak g)] = [\mathfrak g,\gamma_b(\mathfrak g)] = \gamma_{b+1}(\mathfrak g) = \gamma_{1+b}(\mathfrak g). \end{align*} Thus the desired containment holds for $a=1$ for every $b$. Now assume $a \geq 2$ and suppose the estimate is known with first index $a-1$ and arbitrary second index. More explicitly, for every integer $c \geq 1$ we assume \begin{align*} [\gamma_{a-1}(\mathfrak g),\gamma_c(\mathfrak g)] \subseteq \gamma_{a-1+c}(\mathfrak g). \end{align*} We must prove \begin{align*} [\gamma_a(\mathfrak g),\gamma_b(\mathfrak g)] \subseteq \gamma_{a+b}(\mathfrak g) \end{align*} for an arbitrary integer $b \geq 1$. The definition \begin{align*} \gamma_a(\mathfrak g)=[\mathfrak g,\gamma_{a-1}(\mathfrak g)] \end{align*} means that every element of $\gamma_a(\mathfrak g)$ is a $k$-linear combination of brackets $[x,u]$ with $x \in \mathfrak g$ and $u \in \gamma_{a-1}(\mathfrak g)$. Since the Lie bracket is bilinear, it is enough to prove the containment for generators of the bracket subspace. So take \begin{align*} x \in \mathfrak g, \qquad u \in \gamma_{a-1}(\mathfrak g), \qquad v \in \gamma_b(\mathfrak g). \end{align*} We will prove that $[[x,u],v] \in \gamma_{a+b}(\mathfrak g)$. The Jacobi identity rewrites this bracket as a difference of two brackets whose depths can be controlled separately: \begin{align*} [[x,u],v]=[x,[u,v]]-[u,[x,v]]. \end{align*} For the first term, apply the induction hypothesis with $c=b$. Since $u \in \gamma_{a-1}(\mathfrak g)$ and $v \in \gamma_b(\mathfrak g)$, we get \begin{align*} [u,v] \in [\gamma_{a-1}(\mathfrak g),\gamma_b(\mathfrak g)] \subseteq \gamma_{a+b-1}(\mathfrak g). \end{align*} Bracketing with $x \in \mathfrak g$ then gives, by the recursive definition of the lower central series, \begin{align*} [x,[u,v]] \in [\mathfrak g,\gamma_{a+b-1}(\mathfrak g)] = \gamma_{a+b}(\mathfrak g). \end{align*} For the second term, first use the definition of the lower central series to locate the inner bracket: \begin{align*} [x,v] \in [\gamma_1(\mathfrak g),\gamma_b(\mathfrak g)] = [\mathfrak g,\gamma_b(\mathfrak g)] = \gamma_{b+1}(\mathfrak g). \end{align*} Now apply the same induction hypothesis with $c=b+1$. This is valid because the induction is on the first index $a$, not on the sum $a+b$; the first index has decreased from $a$ to $a-1$. Therefore \begin{align*} [u,[x,v]] \in [\gamma_{a-1}(\mathfrak g),\gamma_{b+1}(\mathfrak g)] \subseteq \gamma_{a+b}(\mathfrak g). \end{align*} Both terms in the Jacobi decomposition lie in the $k$-linear subspace $\gamma_{a+b}(\mathfrak g)$, so their difference also lies there: \begin{align*} [[x,u],v] = [x,[u,v]]-[u,[x,v]] \in \gamma_{a+b}(\mathfrak g). \end{align*} By bilinearity and the spanning definition of $[\gamma_a(\mathfrak g),\gamma_b(\mathfrak g)]$, every element of that bracket subspace lies in $\gamma_{a+b}(\mathfrak g)$. This completes the induction on $a$ and proves the estimate for all integers $a,b \geq 1$.[/guided]

[step:Embed each derived term into an exponentially deep lower central term]We prove by induction on $r \geq 0$ that \begin{align*} \mathfrak g^{(r)} \subseteq \gamma_{2^r}(\mathfrak g). \end{align*} For $r=0$, this is equality: \begin{align*} \mathfrak g^{(0)}=\mathfrak g=\gamma_1(\mathfrak g)=\gamma_{2^0}(\mathfrak g). \end{align*} Assume that $\mathfrak g^{(r)} \subseteq \gamma_{2^r}(\mathfrak g)$ for some $r \geq 0$. Taking brackets of subspaces preserves inclusion, so \begin{align*} \mathfrak g^{(r+1)} = [\mathfrak g^{(r)},\mathfrak g^{(r)}] \subseteq [\gamma_{2^r}(\mathfrak g),\gamma_{2^r}(\mathfrak g)]. \end{align*} Applying the lower central bracket estimate with $a=b=2^r$ gives \begin{align*} [\gamma_{2^r}(\mathfrak g),\gamma_{2^r}(\mathfrak g)] \subseteq \gamma_{2^r+2^r}(\mathfrak g) = \gamma_{2^{r+1}}(\mathfrak g). \end{align*} Hence \begin{align*} \mathfrak g^{(r+1)} \subseteq \gamma_{2^{r+1}}(\mathfrak g). \end{align*} The induction is complete.[/step]

[guided]We now compare the two series. The derived series takes brackets of a term with itself, while the lower central series takes brackets with the whole algebra. The lower central bracket estimate shows that self-bracketing a term deep in the lower central series pushes it twice as deep. We prove \begin{align*} \mathfrak g^{(r)} \subseteq \gamma_{2^r}(\mathfrak g) \end{align*} for every $r \geq 0$ by induction. For $r=0$, both sides are $\mathfrak g$: \begin{align*} \mathfrak g^{(0)} = \mathfrak g = \gamma_1(\mathfrak g) = \gamma_{2^0}(\mathfrak g). \end{align*} Assume the containment holds at level $r$. Then every element of $\mathfrak g^{(r)}$ lies in $\gamma_{2^r}(\mathfrak g)$. Since the bracket operation is bilinear, bracketing smaller subspaces gives a smaller bracket subspace: \begin{align*} [\mathfrak g^{(r)},\mathfrak g^{(r)}] \subseteq [\gamma_{2^r}(\mathfrak g),\gamma_{2^r}(\mathfrak g)]. \end{align*} Using the definition of the derived series on the left gives \begin{align*} \mathfrak g^{(r+1)} \subseteq [\gamma_{2^r}(\mathfrak g),\gamma_{2^r}(\mathfrak g)]. \end{align*} Now apply the lower central bracket estimate with $a=b=2^r$: \begin{align*} [\gamma_{2^r}(\mathfrak g),\gamma_{2^r}(\mathfrak g)] \subseteq \gamma_{2^r+2^r}(\mathfrak g) = \gamma_{2^{r+1}}(\mathfrak g). \end{align*} Combining the two containments gives \begin{align*} \mathfrak g^{(r+1)} \subseteq \gamma_{2^{r+1}}(\mathfrak g). \end{align*} Thus the induction proves the claimed exponential containment for all $r \geq 0$.[/guided]

[step:Show that the lower central series is descending] We prove by induction on $m \geq 1$ that $\gamma_m(\mathfrak g)$ is an ideal of $\mathfrak g$. The term $\gamma_1(\mathfrak g)=\mathfrak g$ is an ideal of itself. If $\gamma_m(\mathfrak g)$ is an ideal of $\mathfrak g$, then \begin{align*} \gamma_{m+1}(\mathfrak g) = [\mathfrak g,\gamma_m(\mathfrak g)] \subseteq \gamma_m(\mathfrak g). \end{align*} Moreover, \begin{align*} [\mathfrak g,\gamma_{m+1}(\mathfrak g)] = \gamma_{m+2}(\mathfrak g), \end{align*} so $\gamma_{m+1}(\mathfrak g)$ is also an ideal of $\mathfrak g$. Hence every $\gamma_m(\mathfrak g)$ is an ideal, and the displayed containment shows that the lower central series is descending: \begin{align*} \gamma_{m+1}(\mathfrak g) \subseteq \gamma_m(\mathfrak g) \qquad\text{for every } m \geq 1. \end{align*} [/step]

[step:Use nilpotence to force the derived series to vanish] Since $\mathfrak g$ is nilpotent, there exists an integer $N \geq 1$ such that \begin{align*} \gamma_N(\mathfrak g)=0. \end{align*} Choose an integer $R \geq 0$ such that $2^R \geq N$. The lower central series is descending, so \begin{align*} \gamma_{2^R}(\mathfrak g) \subseteq \gamma_N(\mathfrak g)=0. \end{align*} By the containment proved above, \begin{align*} \mathfrak g^{(R)} \subseteq \gamma_{2^R}(\mathfrak g) = 0. \end{align*} Therefore $\mathfrak g^{(R)}=0$. Hence the derived series terminates at zero, so $\mathfrak g$ is solvable. [/step]