Q:

Showing that a sequence $a_n$ is a solution of the recurrence relation

I'm having some trouble with showing that a sequence $a_n$ is a solution to the recurrence relation $a_n = -3a_{n-1} + 4a_{n-2}$. (See image below). The sequence $a_n$ that is given $= (-4)^n$ . 
I'm given the answer in the solutions manual, but I have absolutely no clue what is going on between step II and III. How did they get rid of the $n-1$ exponent? What did they multiply/divide/subtract/add to the equation?

A:

Since 
$$n-1=1+(n-2)$$
they have
$$\begin{align}-3(-4)^{n-1}+4(-4)^{n-2}&=-3\cdot (-4)^{1+(n-2)}+4(-4)^{n-2}\\&=-3\cdot(-4)^1\cdot (-4)^{n-2}+4(-4)^{n-2}\\&=(-4)^{n-2}(-3\cdot (-4)+4).\end{align}$$

