I'm going through the general method of how you construct the Jordan-normal form of a matrix $A$ and my notes say that $$J_{\lambda ,l} -\lambda I_l : e_l \mapsto e_{l-1} \ldots e_2 \mapsto e_1 \mapsto 0$$ where $J_{\lambda ,l}$ is the elementary Jordan block of the eigenvalue $\lambda$, $I_l$ is the identity matrix of size $l\times l$ and $e_1 ,\ldots ,e_l$ are the vectors of the standard basis for $\mathbb{C}^l$.
Why do these basis vectors map this way?
As
$$J_{\lambda,l} = \begin{bmatrix} \lambda & 1 & 0 & 0 & \cdots & 0 & 0 & 0\\ 0 & \lambda & 1 & 0 & \cdots & 0 & 0 & 0\\ 0 & 0 & \lambda & 1 & \cdots & 0 & 0 & 0\\ \vdots & \vdots & \vdots & \ddots & \ddots & \vdots & \vdots & \vdots\\ 0 & 0 & 0 & \cdots & \lambda & 1 & 0 & 0\\ 0 & 0 & 0 & \cdots & 0 & \lambda & 1 & 0\\ 0 & 0 & 0 & \cdots & 0 & 0 & \lambda & 1\\ 0 & 0 & 0 & \cdots & 0 & 0 & 0 & \lambda \end{bmatrix},$$
we see that
$$J_{\lambda,l} - \lambda I_l = \begin{bmatrix} 0 & 1 & 0 & 0 & \cdots & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & \cdots & 0 & 0 & 0\\ 0 & 0 & 0 & 1 & \cdots & 0 & 0 & 0\\ \vdots & \vdots & \vdots & \ddots & \ddots & \vdots & \vdots & \vdots\\ 0 & 0 & 0 & \cdots & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & \cdots & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & \cdots & 0 & 0 & 0 & 1\\ 0 & 0 & 0 & \cdots & 0 & 0 & 0 & 0 \end{bmatrix}.$$
Note that
$$(J_{\lambda,l} - \lambda I_l)x = \begin{bmatrix} 0 & 1 & 0 & 0 & \cdots & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & \cdots & 0 & 0 & 0\\ 0 & 0 & 0 & 1 & \cdots & 0 & 0 & 0\\ \vdots & \vdots & \vdots & \ddots & \ddots & \vdots & \vdots & \vdots\\ 0 & 0 & 0 & \cdots & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & \cdots & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & \cdots & 0 & 0 & 0 & 1\\ 0 & 0 & 0 & \cdots & 0 & 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ x_3\\ \vdots\\ x_{l-3}\\ x_{l-2}\\ x_{l-1}\\ x_l \end{bmatrix} = \begin{bmatrix} x_2\\ x_3\\ x_4\\ \vdots\\ x_{l-2}\\ x_{l-1}\\ x_l\\ 0 \end{bmatrix}.$$
Therefore, for $2 \leq i \leq l$, $(J_{\lambda, l} - \lambda I_l)e_l = e_{l-1}$ and $(J_{\lambda, l} - \lambda I_l)e_1 = 0$.