In a matrix Lie group $G$, we say that $g\in G$ is unipotent if $$(g-I)^n=0 $$ for some $n\in \mathbb{N}.$
I read in a Tao's article, that
More generally, we say that an element g of a Lie group $G$ is unipotent if its adjoint action $x \mapsto gxg^{-1}$ on the Lie algebra $\mathfrak{g}$ is unipotent.
How can I show that in the matrix Lie group case these definition coincides, i.e., if $(g-I)^n=0$, then $(Ad_g(x)-x)^k =0$ for some $k\in \mathbb{N}$ and $x$ in the Lie algebra of $G$?
Is this even true?