It is clear that $\ker(T - \lambda_i \cdot \mathrm{id}) \subseteq \ker(T - \lambda_i \cdot \mathrm{id})^2$. For the other direction, let $v \in V$ and decompose it as $v = v_1 + \ldots + v_k$ where $Tv_j = \lambda_j v_j$. If $v \in \mathrm{ker}(T - \lambda_i \cdot \mathrm{id})^2$ then
$$ (T - \lambda_i \cdot \mathrm{id})^2(v) = (T - \lambda_i \cdot \mathrm{id}) \left( \sum_{j \neq i} (\lambda_j - \lambda_i) v_j \right) = \sum_{j \neq i} (\lambda_j - \lambda_i)^2 v_j = \sum_{\{j \neq i \, | v_j \neq 0\}} (\lambda_j - \lambda_i)^2 v_j = 0.$$
Since the set $\{ v_j \, | \, v_j \neq 0 \}$ is linearly independent (for it consists of eigenvectors of $T$ that correspond to different eigenvalues) and since $\lambda _i \neq \lambda_j \neq 0$ for all $j \neq i$, the set $\{ v_j \, | \, v_j \neq 0 \}$ must be empty, showing that $v = v_i$ and hence $v \in \mathrm{ker}(T - \lambda_i \cdot \mathrm{id})$.
Thus, $\ker(T - \lambda_i \cdot \mathrm{id}) = \ker(T - \lambda_i \cdot \mathrm{id})^2$ and hence
$$\dim \ker(T - \lambda_i \cdot \mathrm{id}) = \dim \ker(T - \lambda_i \cdot \mathrm{id})^2 \implies \mathrm{rank}(T - \lambda_i \cdot \mathrm{id}) = \mathrm{rank} (T - \lambda_i \cdot \mathrm{id})^2. $$
