Let $N \in \mathcal{M}_{n}(\mathbb{C})$ be a nilpotent matrix of indice $m$ and $A \in \mathcal{M}_{n}(\mathbb{C})$ a matrix such that $A N=N A$. We assume that $A$ is singular. Expressing $(A+N)^{k}$ for $k \in \mathbb{N}$, prove that $\det(A+N) = 0$.
I don't know how to prove this using binomial formula, i can't pursue because of the $A^{k}$ in the formula and the non linearity of the determinant for matrix. I found an explanation in a previous topic but they used the trace and the newton's identity to prove that the characteristic polynomial of A+N and A is the same. What i have not understood is how to apply the newton's identities and the trace.
Pick any nonzero vector $x\in\ker(A)$. Then \begin{aligned} (A+N)^nx &=\sum_{k=0}^n\binom{n}{k}N^{n-k}A^kx\quad\text{(because $AN=NA$)}\\ &=N^nx\quad\text{(because $Ax=0$)}\\ &=0.\quad\text{(because $N^n=0$)}. \end{aligned} Therefore $(A+N)^n$ is singular. In turn, $A+N$ is singular.