Prove that $A+I$ is invertible if $A$ is nilpotent [duplicate]

You can easily prove that if $A^n=0$: $$\left(A+I\right)\left(I-A+A^2-...+(-1)^n A^{n-1}\right) = I +(-1)^{n-1} A^n = I$$ Thus proving that $A+I$ is invertible for any nilpotent $A$.

More generally: A (square) matrix $A$ is invertible if and only if $\lambda=0$ is not an eigenvalue.

Independently of this, we have that if $\lambda$ is an eigenvalue of $A$, then $\lambda+\mu$ is an eigenvalue of $A+\mu I$: if $\mathbf{x}$ is an eigenvector of $A$ corresponding $\lambda$, then $(A+\mu I)\mathbf{x} = A\mathbf{x}+\mu I\mathbf{x} = \lambda\mathbf{x}+\mu\mathbf{x} = (\lambda+\mu)\mathbf{x}$.

Also independently of all of this, if $\lambda$ is an eigenvalue of $A$, then $\lambda^n$ is an eigenvalue of $A^n$. (Careful, though, this one is not reversible: a rotation of $90^{\circ}$ of the plane has no eigenvalues over $\mathbb{R}$, but its square has eigenvalue $-1$. The square of the identity has $(-1)^2$ as an eigenvalue, but the identity does not have $-1$ as an eigenvalue).

So: $$\begin{align*} A+I\text{ is invertible}&\iff 0\text{ is not an eigenvalue of }A+I\\ &\iff -1\text{ is not an eigenvalue of }A. \end{align*}$$ And if $A^n=0$ for some $n\gt 0$, then $-1$ is not an eigenvalue of $A$.