I want to show that for a complex Lie algebra $\mathfrak{g}$ which is solvable that $\mathfrak{g}'$ is nilpotent. Where $\mathfrak{g}'$ is the derived algebra $[\mathfrak{g},\mathfrak{g}]$.
I have shown the converse, since $\mathfrak{g}'$ being nilpotent means that $\mathfrak{g}'$ is solvable and hence the derived series of $\mathfrak{g}'$ terminates, and hence the derived series of $\mathfrak{g}$ (which is the same except for the first step) terminates.
In regard to when $\mathfrak{g}$ is solvable, we have a terminating derived series, but I can't see how this means that the lower central series terminates.
We have $\mathfrak{g}^{(0)}=\mathfrak{g},\quad \mathfrak{g}^{(n+1)}=[\mathfrak{g}^{(n)},\mathfrak{g}^{(n)}]$ terminating, and apparently this means that:
$\mathfrak{g}^0=\mathfrak{g}^{(1)},\quad\mathfrak{g}^{n+1}=[\mathfrak{g}^{(1)},g^n]$, but I don't see it.
Feel free to use whatever notation you want, rather than mine
I think this is only true if $\mathfrak{g}$ is finite-dimensional. Then by Lie's theorem, $\mathfrak{g}$ is a subalgebra of the Lie algebra of upper triangular matrices, so $[\mathfrak{g}, \mathfrak{g}]$ is a subalgebra of the Lie algebra of strictly upper triangular matrices.