Calculating the exponential of a $4 \times 4$ matrix 
Find $e^{At}$, where $$A = \begin{bmatrix} 1 & -1 & 1 &  0\\ 1 &  1 & 0 &  1\\ 0 &  0 & 1 & -1\\ 0 &  0 & 1 &  1\\ \end{bmatrix}$$


So, let me just find $e^{A}$ for now and I can generalize later. I notice right away that I can write
$$A = \begin{bmatrix} B & I_{2} \\ 0_{22} & B \end{bmatrix}$$
where
$$B = \begin{bmatrix} 1 & -1\\ 1 &  1\\ \end{bmatrix}$$
I'm sort of making up a method here and I hope it works.  Can someone tell me if this is correct?
I write:
$$A = \mathrm{diag}(B,B) + \begin{bmatrix}0_{22} & I_{2}\\ 0_{22} & 0_{22}\end{bmatrix}$$
Call $S = \mathrm{diag}(B,B)$, and $N = \begin{bmatrix}0_{22} & I_{2}\\ 0_{22} & 0_{22}\end{bmatrix}$. I note that $N^2$ is $0_{44}$, so
$$e^{N} = \frac{N^{0}}{0!} + \frac{N}{1!} + \frac{N^2}{2!} + \cdots = I_{4} + N + 0_{44} + \cdots = I_{4} + N$$
and that $e^{S} = \mathrm{diag}(e^{B}, e^{B})$ and compute:
$$e^{A} = e^{S + N} = e^{S}e^{N} = \mathrm{diag}(e^{B}, e^{B})\cdot[I_{4}  + N]$$
This reduces the problem to finding $e^B$, which is much easier.
Is my logic correct?  I just started writing everything as a block matrix and proceeded as if nothing about the process of finding the exponential of a matrix would change. But I don't really know the theory behind this I'm just guessing how it would work.

Consider $M(t) = \exp(t A)$, and as you noticed, it has block-diagonal form 
$$
M(t) = \left(\begin{array}{cc} \exp(t B) & n(t) \\ 0_{2 \times 2} & \exp(t B) \end{array} \right).
$$
Notice that $M^\prime(t) = A \cdot M(t)$, and this results in a the following differential equation for $n(t)$ matrix:
$$
   n^\prime(t) = \mathbb{I}_{2 \times 2}  \cdot \exp(t B) + B \cdot n(t)
$$
which translates into 
$$
  \frac{\mathrm{d}}{\mathrm{d} t} \left( \exp(-t B) n(t) \right) = \mathbb{I}_{2 \times 2}
$$
which is to say that $n(t) = t \exp(t B)$.
