What is the Z-transform of $0.8^{n+2}u(n-1)$? I have 2 signals. One is $x(n)=(-0.5)^nu(n)$ and the other one is $y(n)=0.8^{n+2}u(n-1)$. I know that for the first one it is $X(z)= 1/(1+0.5z^{-1})$, but what about the other one? I know $y(n)$ is time shifted but i don't know how to find this z-transform. Any help would be appreciated!
HINT:
If you rewrite $y[n]$ as
$$y[n]=(0.8)^3(0.8)^{n-1}u[n-1]\tag{1}$$
does it become easier to find the $\mathcal{Z}$-transform?