For an ideal $I$ of a Lie algebra $L$ we define $I^{0}=I$ and $I^{k}=[I,I^{k-1}]$ for $k\geq1.$ I want to prove the following
Proposition: If $I$ and $J$ are two ideals of L then $(I+J)^{2n}\subset I^{n}+J^{n}.$
I know this theorem in case of commutative associative rings, but it uses the binomial formula $(x+y)^{k}=\sum\begin{pmatrix}k\\i\end{pmatrix}x^{k-i}y^{i}.$ Then one argues that if $x\in I,$ $y\in J$ and $k=2n,$ then always either $i\geq n$ or $k-i\geq n$ and therefore each elements of the righthand side is either in $I^{n}$ or in $J^{n}.$
Question: is there any generalisation of the binomial formula to the case of Lie algebras or at least noncommutative associative rings?
If no then how one proves the first proposition? I was also trying induction but I didn't came up with anything interesting. I will be thankful for any reference or a hint.
The claim is proved in Proposition $6$ on page $25$ in Jacobson's book on Lie algebras. Writing $IJ$ for $[I,J]$ in the Lie algebra case for ideals, we have for Lie algebra ideals $I,J$ \begin{align*} IJ & =JI, \\ (IJ)K & \subseteq (IK)J+(JK)I \end{align*} by the skewsymmetry and the Jacobi identity. It is straightforwward to show that $(I+J)^m\subseteq I^{[m/2]}+J^{[m/2]}$, see Jacobson.