Let N=beginpmatrix0&1&2\\0&0&3\\0&0&0endpmatrix and M=I+N.

Since $N$ is strictly upper triangular, $N^3=0$ and $N^2$ has only the $(1,3)$ entry, equal to $1\cdot3=3$. Because $I$ and $N$ commute, $M^n=I+nN+\binom n2N^2$. For $n=10$ the $(1,3)$ entry is $10\cdot2+45\cdot3=155$, not $135$, so (A) is false. The $(2,3)$ entry is $10\cdot3=30$, not $20$, so (C) is false. The inverse is $M^{-1}=I-N+N^2$, whose $(1,3)$ entry is $-2+3=1$, so (B) is true. For (D), $M^n=I$ would need $nN+\binom n2N^2=0$, impossible since the $(1,2)$ entry is $n\ne0$, so (D) is false.