For a 2x2 matrix A satisfying $A^k=I$, compute $e^A$ For a 2x2 matrix A satisfying $A^k=I$, compute $e^A$ 
Oh, the exponential of a matrix is: $e^A=\sum_{i=0}^\infty\frac{1}{i!}A^i$

I thought I'd solved the $e^A$ form but I actually did something really silly, now I'm a little stuck.
If $A^k=I$ then $A^{k+1}=A$ and we have a cycle forming, so we will get:
$$\sum^{k-1}_{i=0}\sum^{\infty}_{j=1}\frac{1}{(kj+i)!}A^i$$
(Or something like this form, I don't have paper to hand and just spotted that now)
Is this what the question wants? It doesn't use the 2x2 property.
I believe the answer lies in finding an expression for the inner summation, I can do this for two terms (think of $e+e^{-1}$ all the odd power leave)

If $A^k=I$ then $A$ is full rank.  However there is no necessary relation between the rank $n$ and $k$.
E.g. a $2\times 2$ real orthogonal matrix $A$ can have arbitrarily large multiplicative order (think roots of unity), but also $k \gt 0$ can be the order of an arbitrary size matrix (say $n\times n$).
Matrix exponentiation by diagaonlization 
Among the various ways to evaluate matrix exponential, a diagonal matrix $D = \operatorname{diag}(d_1,\ldots,d_n)$ is an easily computed case:
$$ e^D = \operatorname{diag}(e^{d_1},\ldots,e^{d_n}) $$
Also the exponential function conserves similarity relationships because of the power series definition, so if $A = T D T^{-1}$, then:
$$ e^A = T e^D T^{-1} $$
Not every matrix is diagonalizable, but if $A^k = I$ for $k \gt 0$, then $A$ is diagonalizable.  Because $X^k - 1$ has no repeated roots, the minimal polynomial of $A$ must divide $X^k - 1$ and also have no repeated roots.  From this it follows that $A$ is diagonalizable.
Solutions of $A^k=I$ that are real $2\times 2$ matrices
In particular for the $2\times 2$ real matrices, either the minimum exponent $k$ is one, which implies $A=I$ and $e^I = \operatorname{diag}(e,\ldots,e)$, or the minimum exponent $k$ is greater than one (the multiplicative order of $A$).  In the latter case a $2\times 2$ real matrix $A$ will have a conjugate pair of complex eigenvalues that are primitive $k$th roots of unity:
$$e^{\pm 2\pi im/k} = \cos(2\pi m/k) \pm i\sin(2\pi m/k)$$
where integer $0 \lt m \lt k$ is coprime to $k$.
Then for a suitable diagonalizing similarity transformation $T$:
$$ A = T \pmatrix{e^{2\pi im/k}  & 0 \cr
                       0 & e^{-2\pi im/k} \cr } T^{-1} $$
Per the earlier discussion:
$$ e^A = T \pmatrix{e^{e^{2\pi im/k}}  & 0 \cr
                         0 & e^{e^{-2\pi im/k}} \cr } T^{-1} $$
