TITLE: If $N$ is nilpotent, prove that $\det(I+N)=1$
QUESTION [6 upvotes]: If $N$ is nilpotent, prove that $\det(I+N)=1$.

My attempt:
Since $N$ is nilpotent we have $$\det(N)=0.$$ That is, there exists a non-zero vector $v$ such that $$Nv=0.$$
Now consider $$(I+N)v=Iv+Nv=Iv+0=v.$$
Thus $$(I+N)v=v.$$ That is, $1$ is an eigenvalue of $I+N$. With this, how can we say $\det(I+N)=1$?

REPLY [2 votes]: May be the most simple solution can be - 
Suppose $N$ is the nilpotent matrix.
Then the characteristic polynomial of $N$ is -
$x^k =$det$(xI-N)$
Putting $x=-1$ we get
$(-1)^k = (-1)^k$ det$(I+N)$ 
Hence det$(I+N)$=$1$