The determinant (and also trace) of a nilpotent matrix is always zero, so a nilpotent matrix cannot be invertible. However, if $N$ is nilpotent of index $m$, then $I+N$ and $I-N$ are invertible and the inverses are given by
\begin{align*} (I+N)^{-1}&=\sum_{k=0}^{m-1}(-N)^k=I-N+N^2-N^3+\cdots+(-N)^{m-1}\\ (I-N)^{-1}&=\sum_{k=0}^{m-1}N^k=I+N+N^2+N^3+\cdots+N^{m-1} \end{align*}
We have shown here that invertible operators have square roots. By the same token, we see that the square roots of $(I+N)^{-1}$ and $(I-N)^{-1}$ exist. But how do we find them? Let us denote them by $(I+N)^{-\frac{1}{2}}$ and $(I-N)^{-\frac{1}{2}}$, respectively. Then by the same manner we proved the existence of $\sqrt{N+1}$ here, we obtain
\begin{align*} (I+N)^{-\frac{1}{2}}&=I-\frac{1}{2}N-\frac{3}{8}N^2-\frac{11}{16}N^3+\cdots\\ (I-N)^{-\frac{1}{2}}&=I+\frac{1}{2}N+\frac{3}{8}N^2+\frac{5}{16}N^3+\cdots \end{align*}