I was reading this book and on page 76 the author defines $\mathcal{N}(\mathbf{g})$ to be the set of all elements of $\mathbf{g}$ that are in the generalized eigenspace of some other element's adjoint, of nonzero eigenvalue. The author then tells us to "Notice that every element of $\mathcal{N}(\mathbf{g})$ is ad nilpotent..."
I do not notice it, and I find it hard to prove even in the true eigenvector case. This involves noticing that if $[y, x]=x$ then $[\text{ad}y, P(\text{ad}x)]=\text{ad}xP'(\text{ad}x)$ in the adjoint representation whatever polynomial $P$ is, and then letting $P$ be the minimal polynomial of $\text{ad}x$ and looking at degrees proves that $zP'(z)=n\cdot P(z)$, and then the only possible root is $0$.
Is there an easy way to see this in general? Even a case like $(\text{ad}y-1)^2x=0$ seems intractible.
Because for $x\in\mathfrak{g}_\alpha$, $\mathrm{ad}(x)$ maps $\mathfrak{g}_\beta$ into $\mathfrak{g}_{\beta+\alpha}$. So $\mathrm{ad}(x)^k$ maps $\mathfrak{g}_\beta$ into $\mathfrak{g}_{\beta+k\alpha}$. Assuming that the characteristic is zero (this is somewhat hidden in your book), this is zero for some $k\le\dim(\mathfrak{g})$. Hence $\mathrm{ad}(x)^{\dim(\mathfrak{g})}=0$.