**Part 1**

Set $B = I - A$ so that $A = I - B$. We have $A^2 = (I - B)^2 = I - 2B + B^2 = 0$ so that $I = B(B-2I)$. We must now check that $|B| \neq 0$. Taking the determinant of both sides, we get $1 = |B||B - 2I|$ so regardless of the value of $|B - 2I|$, we have that $|B| \neq 0$ so the matrix is invertible.

**Parts 2 and 3**

The idea for these parts follows naturally from the proof for $n = 2$. Since we have \[ \underbrace{(I - B)(I - B) \cdots (I - B)}_{\text{n times}} = 0, \]

we can multiply the $I$ terms and obtain \[ -B\Bigl[ (-1)^n B^{n-1} + \ldots \Bigr] = I. \]

Reasoning in the same way, we get $|B| \neq 0$ so that $B = I - A$ is invertible.

**Part 4**

Just factor. We have \[ A^2 + 2A + I = (A + I)^2 = 0 \]

Set $C = A + I$ so that $C^2 = 0$, and note that $I - C = -A$ must be invertible by the first part. Thus $A$ must be invertible.