$a_{n-1} = ca_{n-2} $
Hence
$a_n = c \cdot c \cdot a_{n-2} $
$ = c \cdot c \cdot c \cdot a_{n-3} $
......
$ = c^na_0 $
Why is there a iteration on the constant $c$ ?
$a_{n-1} = ca_{n-2} $
Hence
$a_n = c \cdot c \cdot a_{n-2} $
$ = c \cdot c \cdot c \cdot a_{n-3} $
......
$ = c^na_0 $
Why is there a iteration on the constant $c$ ?
Your basic equation says that to get from $a_{n-2}$ to $a_{n-1}$ you multiply by $c$. If one step corresponds to multiplying by $c$, two steps should be multiplying by $c$ twice, which is $c\cdot c=c^2$ and $n$ steps should be multiplying by $c^n$.
An explicit example: take $a_0=1, c=2.$ Then your first equation says (taking $n=1$) $a_1=2a_0=2$, Then (taking $n=2$) $a_2=2a_1=4$ Does this help?