If $N$ is nilpotent, prove that $\det(I+N)=1$ 
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 $\det(I+N)=1$?
 A: Assuming we're talking about matrices (i.e. in finite dimensions), then $N$ is nilpotent if and only if $0$ is the only eigenvalue of $N$. Adding $I$ to a matrix adds $1$ to all of its eigenvalues, so $I + N$ has only the eigenvalue $1$. The determinant, being the product of its eigenvalues (up to multiplicity), is therefore $1$.
We can easily prove it by contradiction. Suppose $I + N$ has a determinant not equal to $1$. Then, at least one of the eigenvalues must not equal $1$, so there exists a $\lambda \neq 1$ and non-zero $v$ such that,
$$\lambda v = (I + N)v = v + Nv \implies Nv = (\lambda - 1)v.$$
Since $\lambda \neq 1$ and $v \neq 0$, the resulting vector $(\lambda - 1)v \neq 0$, and is still an eigenvector for $N$ corresponding to $\lambda - 1$. Therefore, by induction,
$$N^k v = (\lambda - 1)^k v \neq 0,$$
which contradicts $N$ being nilpotent.
A: If $N^k = 0$ and $X = I + N$ then $(X - I)^k = 0$. So $X$ satisfies the polynomial $(x - 1)^k$. This means that every eigenvalue of $X$ is $1$.
To see this, let $v$ be an eigenvector with eigenvalue $\lambda$. Since $(X - I)^k = 0$ we know that $(X - I)^kv = 0$. If we expand the left-hand side using the Binomial Theorem we get
$$ \sum_{r = 0}^k \binom{k}{r}X^r(-I)^{k-r}v = \sum_{r = 0}^k \binom{k}{r}(-1)^{k-r}X^rv = \sum_{r = 0}^k \binom{k}{r}(-1)^{k-r}\lambda^rv = (\lambda - 1)^kv. $$
Thus $(\lambda - 1)^kv = 0$ and hence $\lambda = 1$.
Since $\det X$ is the product of the eigenvalues of $X$ (with multiplicity) it follows that $\det X = 1$.
A: 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$(I+N)$=$1$
