I read the following result:
If $A$ is $n\times n$ matrix and $\rho(A)$ denotes the spectral radius of $A$. Then if $\rho(A)<1$, then $\lim\limits_{m\rightarrow \infty} A^m\rightarrow \boldsymbol{0}$, where $\boldsymbol{0}$ is zero matrix.
Now result is easily proved to be true for diagonalizable matrices because if $PAP^{-1}$ is diagonal matrix $D$, then $D=diag(d_1, d_2, \dots, d_n)$ where $|d_i|<1$. Then $PA^mP^{-1}=diag(d_1^m, d_2^m, \dots, d_n^m)\rightarrow \boldsymbol{0}$. So $\lim\limits_{m\rightarrow \infty} A^m\rightarrow \boldsymbol{0}$.
Now we know that $\lim\limits_{m\rightarrow \infty} A^m\rightarrow \boldsymbol{0}$ is same as $\lim\limits_{m\rightarrow \infty} \|A^m\|\rightarrow \boldsymbol{0}$ for any matrix norm $\|\cdot\|$ on $\mathbb M_n(\mathbb C)$.
And in any matrix norm, diagonalizable matrices are dense in $\mathbb M_n(\mathbb C)$. Can we use some continuity argument to prove the result for any matrix? (we also know eigenvalues are continuous function on $\mathbb M_n(\mathbb C)$.