Jordan-block/2/Powers/Induction/Exercise/Solution

For

${\displaystyle {}n=0,1\,,}$

the statement is clear. Suppose now that the statement is already proven for some ${\displaystyle {}n}$. Then

${\displaystyle {}{\begin{aligned}M^{n+1}&=M^{n}\circ M\\&={\begin{pmatrix}\lambda &1\\0&\lambda \end{pmatrix}}^{n}\circ {\begin{pmatrix}\lambda &1\\0&\lambda \end{pmatrix}}\\&={\begin{pmatrix}\lambda ^{n}&n\lambda ^{n-1}\\0&\lambda ^{n}\end{pmatrix}}\circ {\begin{pmatrix}\lambda &1\\0&\lambda \end{pmatrix}}\\&={\begin{pmatrix}\lambda ^{n+1}&\lambda ^{n}+n\lambda ^{n-1}\cdot \lambda \\0&\lambda ^{n+1}\end{pmatrix}}\\&={\begin{pmatrix}\lambda ^{n+1}&(n+1)\lambda ^{n}\\0&\lambda ^{n+1}\end{pmatrix}}.\end{aligned}}}$