A basis-shift-down operator is nilpotent of index $n$

Let $T:\mathbb{R}^n\to\mathbb{R}^n$ be defined as $T(\bar e_1)=\bar0,\ T(\bar e_j)=\bar e_{j-1}, j=2,\dots,n$ where $\{\bar e_1,\bar e_2,\dots,\bar e_n\}$ is the standard basis of $\mathbb{R}^n$. Then: