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?

A very elegant, and a bit more advanced way of showing it is the following. Consider the Initial Value Problem
$$
\left\{\begin{array}{cc}
X'=MX, \\ X(0)=I,
\end{array}
\right.
$$
where $X$ is the unknown, an $N\times N$ matrix and $I$ the identity matrix. Picard-Lindelöf Theorem, for linear systems, implies that the recursive sequence 
$$
X_{n+1}(t)=I+\int_0^t MX_n(s)\,ds,
$$
converges uniformly in every closed interval. It is readily seen that
$$
X_n=\sum_{k=0}^n\frac{t^kM^k}{k!}.
$$
Hence, the Neuman series $\sum_{n=0}^\infty\frac{t^nM^n}{n!}=\exp(tM)$ is has infinite radius of convergence and satisfies the equation
$$
\frac{d}{dt}\exp(tM)=M\exp(tM).
$$
