$ L$ is semisimple Lie algebra in ${\rm gl}\ V$ where $V$ is a complex vector space. Then we have two Jordan forms $$ x=d+n,\ {\rm ad}_x = {\rm ad}_d + {\rm ad}_n\ (x\in L) $$
(cf. Theorem 9.15 in the Erdmann and Wildon's book. The following proof is in page 87.)
Sketch of Proof We have Jordan : $$ {\rm ad}_x =\sigma + \nu $$ where $\sigma$ is diagonalizable and $\nu$ is nilpotent.
Since $L$ is semisipmle we have $$\sigma={\rm ad}_d,\ \nu = {\rm ad}_n $$ where $d,\ n\in L$ (cf. proposition 9.14)
So we must prove that $$ x= d+n $$ is Jordan.
(Commutativeness) $$ {\rm ad}_{[d,n]} =[{\rm ad}_d,{\rm ad}_n]=0$$ Since $L$ is semisimple so $ {\rm ad} $ is an Lie algebra isomorphism and $ [d,n]=0$
(Nilpotency of $n$) Note that there exists $p$ s.t. $${\rm ad}_n = p({\rm ad}_x) = \sum_{i=0}^k c_i ({\rm ad}_x)^i$$ so that $$0=({\rm ad}_n)^K(x)= (c_0)^K (x),\ c_0=0$$
But i cannot proceed anymore
Question : I have two questions : I cannot prove that $n$ is nilpotent and $d$ is diagonalizable.
Thank you