If $A\in M_n(R)$ and $A$ is supertriangular then $A^n=0$

theorem: Let $R$ be commutative ring with identity.If an n-square matrix $A\in Ma{t}_n(R)$is supertriangular then $A^n=0$.

proof: Find the characteristic polynomial of $A$ by computing the determinant of $A-tI$. The square matrix $A-tI$ is a triangular matrix. The determinant of a triangular matrix is the product of the diagonal element of the matrix. Therefore the characteristic polynomial is $p(t)=t^n$ and by the Cayley-Hamilton theorem the matrix $A$ satisfies the polynomial. That is $A^n=0$.
QED