$I+B$ is invertible if $B^{k} = 0$ If $B$ is nilpotent and $B^{k} = 0$ (and B is square), how should I go around proving that $I + B $ is invertible? I tried searching for a formula - $I = (I + B^{k}) = (I + B)(???)$
But I didn't get anywhere :(

Factor $I-B^k$ and $I-B$ is one factor, but it also equals $I$ since $B^k=0$.
