find exponential of 4x4 matrix Hi i am trying to understand  ,  consider the matrix : 
$M = \begin{pmatrix}
0 &-2  & 4 &-2 \\ 
1 &1  &-2  &-1 \\ 
0 &0  &0  &0 \\ 
1 &-1  &2  &-3 
\end{pmatrix}  $ 
$M^2 = \begin{pmatrix}
-4 &0  & 0 &8 \\ 
0 &0  &0  &0 \\ 
0 &0  &0  &0 \\ 
-4 &0  &0  &8 
\end{pmatrix} ,\qquad
$
$
M^3 = \begin{pmatrix}
8 &0  & 0 &-16 \\ 
0 &0  &0  &0 \\ 
0 &0  &0  &0 \\ 
8 &0  &0  &-16 
\end{pmatrix}  $ then $M^n = 2^{n-2}M^2$ 
how to determine $e^M$ ? i known just that $e^M = \sum_{n=0}^{+\infty}\frac{1}{n!}M^n $ but i don't know determine

We have :
$$e^M = I + M + M^2 \cdot\sum_{i=2}^{\infty} \frac{2^{i-2}}{i!}$$
Using the fact that : $e^x = \sum_{i=0}^{\infty} \frac{x^i}{i!}$
We get : 
$$ \sum_{i=2}^{\infty} \frac{2^{i-2}}{i!} = \frac{1}{4} \cdot \sum_{i=2}^{\infty} \frac{2^i}{i!} = \frac{1}{4} \cdot ( e^2-3)$$
So we have :
$$e^M = I + M + \frac{e^2-3}{4} \cdot M^2$$
