Is there a shortcut for taking this matrix to a certain power? One can take a diagonal matrix to a certain power by just taking diagonal elements that power. There is a similar polynomial-time (in respect to the matrix dimensions) shortcut for triangular matrices.
Is there a shortcut for exponentiating a square matrix of the form
$$
\left( \begin{array}{cccc}
a & b & c & d \\
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \end{array} \right)
$$
, where $a$, $b$, $c$, and $d$ are constants?

By using Cayley Hamilton Theorem. If you are doing the calculations by hand, then the problem is simple if the characteristic polynomial has simple roots. 
The general approach is to divide $x^n$ by $x^4-a x^3 - b x^2 - c x - d$ find the remainder. Thus suppose
$$
x^n = Q(x) (x^4-a x^3 - b x^2 - c x - d) + { \alpha x^3 + \beta x^2 + \gamma x + \delta } \tag 1
$$
Here we do not need the quotient $Q(x)$ but just the remainder $\alpha x^3 + \beta x^2 + \gamma x + \delta $
Then, replacing $x$ by $A$ we have
$$
A^n = Q(A) (A^4-a A^3 - b A^2 - c A- d I) + { \alpha A^3 + \beta A^2 + \gamma A + \delta I } \tag 2
$$
In (2) we have $$(A^4-a A^3 - b A^2 - c A- d I) =0 ~\text{Verify this by direct calculation!}$$
so
$$
A^n = { \alpha A^3 + \beta A^2 + \gamma A + \delta I }$$
One can easily calculate
$$A^2=\pmatrix{b+a^2&c+a\,b&d+a\,c&a\,d\cr a&b&c&d\cr 1&0&0&0\cr 0&1&
 0&0\cr }$$
$$A^3=\pmatrix{c+2\,a\,b+a^3&d+a\,c+b^2+a^2\,b&a\,d+b\,c+a^2\,c&
 \left(b+a^2\right)\,d\cr b+a^2&c+a\,b&d+a\,c&a\,d\cr a&b&c&d\cr 1&0&
 0&0\cr }$$
The key to the answer is finding the remainder. One can use recursion (synthetic division) or remainder theorem to find the remainder.
Here is a simple example on finding the remainder when the characteristic equation has simple roots.
Example: Let
$$a=2 , b=1 , c=-2 , d=0  $$
Then the roots of $x^4-a x^3 - b x^2 - c x - d$ are $x=0$, $x=1$, $x=-1$ and $x=-2$.
Now let
$$x^n = Q(x) (x^4-a x^3 - b x^2 - c x - d) + \alpha x^3 + \beta x^2 + \gamma x + \delta$$
We substitute different roots we got as 
Set $x=0$
$$ 0 = \delta \tag 3$$
Set $x=1$:
$$
1 = \alpha+\beta+\gamma+\delta \tag 4$$
Set $x=-1$
$$
(-1)^n = -\alpha+\beta-\gamma+\delta \tag 5$$
Set $x = 2$
$$
2^n = 8 \alpha + 4 \beta + 2 \gamma + \delta \tag6$$
We can solve to get
$$ \alpha={{2^{n}-\left(-1\right)^{n}-3}\over{6}} , \beta={{
 \left(-1\right)^{n}+1}\over{2}} , \gamma=-{{2^{n}+2\,\left(-1\right)
 ^{n}-6}\over{6}} , \delta=0  $$
So we have
$$
A^n = 
{{2^{n}-\left(-1\right)^{n}-3}\over{6}} A^3 +
{{
 \left(-1\right)^{n}+1}\over{2}} A^2 + {{2^{n}+2\,\left(-1\right)
 ^{n}-6}\over{6}} A$$
