Let $V$ be an $n$-dimensional vector space and let $T$ be a linear operator on $V$. Suppose that there exists some positive integer $k$ so that $T^k=0$. Prove that $T^n=0$.
My attempt: Let $M_T = \{f\in F[x] \mid f(T)=0\}$. Let $m$ be minimal polynomial of $T$. Since $T^k=0$, we have $x^k\in M_T$. So $m|x^k$. Which implies $m=x^i$, for some $1\leq i\leq k$. By Cayley-Hamilton theorem, $\deg (m)\leq n$. So $m=x^i$, for some $1\leq i\leq n$. If $i=n$, then we’re done. If $1\leq i\lt n$, then $x^{n-i}\cdot m=x^{n-i}\cdot x^i=x^n$. Since $m(T)=0$, we have $(x^{n-i}\cdot m)(T)=T^{n-i}\cdot m(T)=0=x^n(T)=T^n$. Thus $T^n=0$. Is my proof correct?
To be honest, I’m not completely satisfied with my following step in proof: $m|x^k$ implies $m=x^i$, for some $1\leq i\leq k$. How to show we can’t write $x^k$ as product of two “wired” polynomial?
