What is wrong with this "proof"? $1$ is always an eigenvalue for $I + A$ ($A$ is nilpotent)? 
Consider the nilpotent matrix $A$ ($A^k = 0$ for some positive $k$).
  It is well known that the only eigenvalue of $A$ is $0$.
Then suppose $\lambda$ is any eigenvalue of $I + A$ such that 
  $(I + A) \mathbf{v} = \lambda \mathbf{v}$ where ($\mathbf{v} \neq \mathbf{0}$).
Then $I \mathbf{v} + A \mathbf{v} = \lambda \mathbf{v} \implies 
 \mathbf{v} + A \mathbf{v} = \lambda \mathbf{v} \implies A \mathbf{v} =
(\lambda - 1) \mathbf{v}$
We know that $\lambda - 1 = 0$ because the only eigenvalue of a
  nilpotent matrix is $0$. Therefore $\lambda = 1$

This "proof" seems to indicate that $1$ is always an eigenvalue for the sum of the identity matrix with any nilpotent matrix, but I believe I have a counterexample that disproves this.
I believe my error was in assuming that $I + A$ has eigenvalues -- but I do not know how I could prove/disprove this.
If someone could help me see where I've gone wrong I would greatly appreciate it!

You have proven that if $A$ is nilpotent, then the eigenvalue of $A+I$ (NOT the eigenvalues of $A$) is equal to $1$. There's nothing wrong with the proof.
