Prove that if matrix $A$ is nilpotent, then $I+A$ is invertible. [duplicate]

So my friend and I are working on this and here is what we have so far.

We want to show that $\exists \, B$ s.t. $(I+A)B = I$. We considered the fact that $I - A^k = I$ for some positive $k$. Now, if $B = (I-A+A^2-A^3+ \cdots -A^{k-1})$, then $(I+A)B = I-A^k = I$. My question is: in matrix $B$, why is the sign for $A^{k-1}$ negative? Couldn't it be positive, in which case we'd get $(I+A)B = I + A^k$?

Thank you.

It's the usual polynomial identity $$ 1 - x^{k} = (1 - x)(1 + x + x^{2} + \dots + x^{k-1}), $$ where you are substituting $x = -A$.

By incrementing if necessary by $1$, we can make sure that $k$ is *odd. Then we have the familiar identity (for odd $k$) $1+x^k=(1+x)(1-x+x^2-x^3+\cdots +x^{k-1})$.

As you can see already from your formula for $B=I-A\mathbf{+} A^2-\dots$ the sign can be $+$ for example if $k=2$. In fact the sign that occurs is $(-1)^{k-1}$.