Exponential of a matrix always converges I am trying to show that the exponential of a matrix converges for any given square matrix of size $n\times n$:
$M\mapsto e^M$ e.g. $\displaystyle e^M = \sum_{n=0}^\infty \frac{M^n}{n!}$
Can I argue that: Since $n!$ necessarily grows faster than $k^n$ will, that this converges. This seems to be an obvious fact, since:
$$n!=1\times 2\times 3\times \cdots \times k\times (k+1)\times (k+2)\times \cdots$$
$$k^n=k\times k\times k \times\cdots\times k \times k\times \cdots$$

If we have some $q\times q$ matrix, with $a$'s in each position(which will grow as fast as we make our $a$ and $q$ large) we still only get increasing at a rate of $q^{n-1}\times a^n$

In light of the comments, I know that in this banach space, I need only show that $\displaystyle e^M = \sum_{n=0}^\infty \frac{||M||^n}{n!}$ converges. Now I have many matrix norms to choose from, and I can't seem to get a good argument going rigorously. Any ideas?

For any nxn matrix A, the sequence
$I+A+\frac{A^2}{2!}+...$
converges.
Proof
Let m be the largest 
$|a_{ij}|$ in A. Then
The biggest element in first term is 1.
The biggest element in second term is m.
The biggest element in third term is $\leq\frac{nm^2}{2!}$.
The biggest element in fourth term is $\leq\frac{n^2m^3}{3!}$. etc.
Any ij sequence is dominated by 1, $m$, $\frac{nm^2}{2!}$, $\frac{n^2m^3}{3!}$......$\frac{n^{k-2}m^{k-1}}{{k-1}!}$.....
Applying the ratio test to this maximal sequence gives
$\frac{n^{k-1}m^{k}}{{k}!} \frac{{(k-1)}!}{n^{k-2}m^{k-1}} = \frac{nm}{k}$
 Since n and m are fixed, the ratio goes to 0 as $k \to \infty $, proving (absolute) convergence.
