One distinct eigenvalue $\iff A-\lambda E_n$ is nilpotent Let $A\in \mathbb{C}^{n\times n}$ and $\lambda \in \mathbb{C}.$ $\lambda$ only eigenvalue of $A \iff A-\lambda E_n$ is nilpotent.
So the first direction: let $\lambda$ be the only eigenvalue of $A$ then the characteristic polynomial of $A$ has the form $\chi_A(\lambda)=\left(x-\lambda\right)^n$ and so does the minimal polynomial $\mu_A(\lambda)=\left(x-\lambda\right)^m, m\le n$ then by Cayley Hamilton theorem, the matrix satisfies $(A-\lambda E_n)^n=0$ and also $(A-\lambda E_n)^m=0$ so $A$ is nilpotent of degree $m.$
Any help with the other implication would be appreciated.

Suppose $(A-\lambda I)^m =0$ and $Av = \mu v$ with $v\neq 0$.
Then 
$(A-\lambda I)v = (\mu-\lambda) v$ and so
$(A-\lambda I)^m = (\mu-\lambda)^m v$. Since $(A-\lambda I)^m =0$
we must have $\mu=\lambda$.
