We have defined the derived series $\{\mathscr{D}^k\mathfrak{g}\}$ inductively by $$\mathscr{D}^1\mathfrak{g}=[\mathfrak{g},\mathfrak{g}],\ \mathscr{D}^k\mathfrak{g}=[\mathscr{D}^{k-1}\mathfrak{g},\mathscr{D}^{k-1}\mathfrak{g}].$$ Now I hope to prove that $[\mathfrak{g},\mathscr{D}^k\mathfrak{g}]=\mathscr{D}^k\mathfrak{g}$, then $\mathscr{D}^k\mathfrak{g}$ is an ideal in $\mathfrak{g}$. It is easy for me to show that $[\mathfrak{g},\mathscr{D}^1\mathfrak{g}]=\mathscr{D}^1\mathfrak{g}$, and that is equivalently $[\mathfrak{g},\mathfrak{g}]$ is an ideal of $\mathfrak{g}$. And then? I find it difficult to prove this question by induction. Or are there any other ways to prove it?
I'm a novice in learning Lie algebras and this question may seem quite easy for you. I'd be extremely grateful if you can give me more details to help me!
If $\mathfrak{h}$ is an ideal of $\mathfrak{g}$, then $[\mathfrak{g},D\mathfrak{h}]\subset [\mathfrak{h},[\mathfrak{g},\mathfrak{h}]]$ by the Jacobi identity, which is contained in $[\mathfrak{h},\mathfrak{h}]=\mathfrak{h}$ since $\mathfrak{h}$ is an ideal. So $D\mathfrak{h}$ is an ideal as well.
It follows by induction on $k$ that $D^k\mathfrak{h}$ is an ideal. This applies in particular to $\mathfrak{h}=\mathfrak{g}$.