Exercise 8.A.9

Proof. We have

where the first equality follows from 8.4 and the third because ${(\mathit{ST})}^{\dim <!--\; FUNCTION\; APPLICATION\; -->V}=0$ (by 8.18). Thus ${(\mathit{TS})}^{\dim <!--\; FUNCTION\; APPLICATION\; -->V}=0$ and so $\mathit{TS}$ is nilpotent. □