Inverse of sum of nilpotent matrix and identity matrix

Let $I=I_n$, let $M = A + \lambda I$, and let $B = M^{-1} - \lambda^{-1}I$.

You want to show $B^{m-1} \ne 0$.

\begin{align*} \text{Then}\;\;&MM^{-1}=I\\[4pt] \implies\;&(A+\lambda I)(B+\lambda^{-1}I)=I\\[4pt] \implies\;&AB+\lambda^{-1}A + \lambda B = 0\\[4pt] \implies\;&B(A + \lambda I) = -\lambda^{-1}A\\[4pt] \implies\;&BM = -\lambda^{-1}A\\[4pt] \implies\;&B=(-\lambda^{-1}A)M^{-1}\\[4pt] \implies\;&B^{m-1}=(-\lambda^{-1}A)^{m-1}M^{-(m-1)} &&\text{[since $M$ commutes with A]}\\[4pt] \implies\;&B^{m-1}\ne 0 &&\text{[since $A^{m-1} \ne 0$]}\\[4pt] \end{align*} Note that $M^{-1}$ has only one distinct eigenvalue, namely $\lambda^{-1}$, hence the characteristic polynomial for $M^{-1}$ is $(x-\lambda^{-1})^n$.

But $M^{-1}$ satisfies $(x-\lambda^{-1})^m=0$, since $B^m=0$, and $M^{-1}$ does not satisfy $(x-\lambda^{-1})^{m-1}=0$, since $B^{m-1} \ne 0$. It follows that the minimal polynomial for $M^{-1}$ is $(x-\lambda^{-1})^m$.