Let $n \in \mathbb N$, $A \in \mathbb C^{n \times n}$ be nilpotent and $l\in \mathbb N$. Further, let $$p = \sum_{i=0}^n \alpha_i A^i \in \mathbb C[t]$$ be a polynomial. Show that zero is the only eigenvalue of $A$. Then calculate $\det(p(A))$.
My approach:
Since A is nilpotent, there is a $k\in N$ with $A^k=0$. Let $\lambda $ a eigenvalue of $A$ with eigenvector $v$. Then, $$0=A^kv=A^{k-1}Av=A^{k-1}\lambda v=A^{k-2}A\lambda v=\lambda A^{k-2}Av=\lambda A^{k-2}\lambda v=\dots=\lambda^kv$$and therefore $\lambda^k=0 \Rightarrow \lambda=0$. Therefore $\lambda=0$ is the only eigenvalue of $A$.
For the second part I thought to use that $A$ is similiar to $S^{-1}JS$ with $S\in Gl_n(\mathbb C^{n,n})$ and $J$ is a jordan block of size $n$. Then, $$\det(p(A))=\det(p(S^{-1}JS)=\det (\alpha_0I+\alpha_1S^{-1}JS+\dots+\alpha_nS^{-1}J^nS)=\det(S^{-1}(\alpha_0I+\alpha_1J+\dots+\alpha_nJ^n)S)=\det(\alpha_0I+\alpha_1J+\dots\alpha_nJ^n)$$ and since $J$ is upper triangular with zeros on its diagonal, we recieve $\det(p(A))=\alpha_0^n$.
Does this argumentation seems to be okay?