Given $\mathbf{J}_t(\lambda)=\begin{pmatrix}\lambda&1&&\\&\lambda&1&{\LARGE\mathbf{0}}\\&{\LARGE\mathbf{0}}&\ddots&\ddots\\&&&\lambda\end{pmatrix}\in\mathcal{M}_{t\times t}(\mathbb{F})$, where $\lambda\in\mathbb{F}$. Suppose $\lambda\ne0$, show that $m(x)=\left(x-\lambda^{-1}\right)^t$ is the minimal polynomial of $\mathbf{J}^{-1}$.
It is clear that $m(x)$ is monic. I need help in showing that $m\left(\mathbf{J}^{-1}\right)=\mathbf{0}$. Any help would be appreciated.
ADDED
Note that $m_\mathbf{J}(\mathbf{J})=\mathbf{0}$, i.e. $\left(\mathbf{J}-\lambda\mathbf{I}\right)^t=\mathbf{0}$. Then \begin{align} m\left(\mathbf{J}^{-1}\right)&=\left(\mathbf{J}^{-1}-\lambda^{-1}\mathbf{I}\right)^t\\&=(-1)^t\left(\lambda^{-1}\mathbf{I}-\mathbf{J}^{-1}\right)^t\\&=(-1)^t\left(\lambda^{-1}\mathbf{JJ}^{-1}-\mathbf{J}^{-1}\right)^t\\&=(-1)^t\left(\mathbf{J}-\lambda\mathbf{I}\right)^t\left(\lambda^{-1}\mathbf{J}^{-1}\right)^t\\&=\mathbf{0} \end{align}
If I want to show that if $p(x)$ is a nonzero polynomial over $\mathbb{F}$ such that $p\left(\mathbf{J}^{-1}\right)=\mathbf{0}$, then the degree of $m(x)$ $\le$ the degree of $p(x)$, is my following argument correct?
Suppose there exists $p(x)=a_0+a_1x+\cdots+a_sx^s$ such that $p\left(\mathbf{J}^{-1}\right)=\mathbf{0}$, then$$a_0+a_1\mathbf{J}^{-1}+\cdots+a_s\left(\mathbf{J}^{-1}\right)^s=\mathbf{0}$$Multiplying both sides by $\mathbf{J}^s$, we have$$a_s+a_{s-1}\mathbf{J}+\cdots+a_0\mathbf{J}^s=\mathbf{0}$$Let $q(x)=a_s+a_{s-1}x+\cdots+a_0x^s$. Since $t$ $=$ degree of $m_\mathbf{J}(x)$ $\le$ degree of $q(x)$ $=$ $s$, we have shown that $t$ = degree of $m(x)$ $\le$ degree of $p(x)$.$\qquad\mathbf{Q.E.D}$
Hint: The minimal polynomial of $\mathbf {J}_t(\lambda)$ is $(x-\lambda)^t$, thus $(\mathbf J-\lambda \mathbf I_{t\times t})^t=\mathbf 0_{t\times t}$. Multiply this equality by $(\lambda \mathbf J)^{-t}$.