How to show that there exist a m such that $a_0I+a_1T+dots+a_mT^m=0$?

Let $V$ be a finite dimensional vector space over a field $mathbb{F}$, with a basis ${ b_1,b_2,cdots,b_m}$. If $T$ is a linear transformation on $V$, then $T$ is completely determined by the unique constants ${ a_{r,s} }$, for $1 le r,s le m$ such that $$ T(b_1) = a_{1,1}b_1+a_{1,2}b_2+cdots+a_{1,m}b_m \ T(b_2) = a_{2,1}b_1+a_{2,2}b_2+cdots+a_{2,m}b_m \ vdots \ T(b_m) = a_{m,1}b_1+a_{m,2}b_2+cdots+a_{m,m}b_m. $$ So $T$ is represented by an $mtimes m$ matrix over $mathbb{F}$. The set of $mtimes m$ matrices over $mathbb{F}$ is an $m^2$ dimensional vector space. Therefore ${ I,T,T^2,cdots,T^{mtimes m}}$ is a linearly-depenedent set of matrices, which gives constants $alpha_0,alpha_1,cdots,alpha_m$ that are not all $0$ such that $$ alpha_0 I + alpha_1 T + cdots + alpha_m T^{m^2} = 0. $$