Using the time shifting, time reversal, and scaling, I want to derive the form of the Z Transform of
$$x[n]=-a^n u[-(n+1)]$$
$u[n]$ is the discrete-time unit step function:
$$ u[n] \triangleq \begin{cases} 0 \qquad & n < 0 \\ 1 \qquad & n \ge 0 \\ \end{cases}$$
(disregarding the ROC)
given the Z transform of the unit step:
$$ U(z) = \mathcal{Z} \Big\{u[n] \Big\} = \frac{1}{1-z^{-1}} $$
Let $X(z)$ be informally, the expected form of the Z Transform that $x[n]$ will be taking:
Using the time reversal property
$$ X(z) = U(z^{-1}) $$
Applying the time shifting property:
$$ X(z) = z^{1}U(z^{-1}) $$
Then, by scaling
$$ X(z) = zU(a^{-1}z^{-1}) $$
Returning to the Z Transform of $x[n]$
$$\begin{align} \mathcal{Z} \Big\{-a^nu[-(n+1)] \Big\} &= -zU(a^{-1}z^{-1}) \\ & \Longrightarrow \frac{-z}{1-az} \\ \end{align}$$
which is not the correct answer
$$ \frac{1}{1-az^{-1}} $$
Please note that I momentarily disregarded the concern for ROC in order to know if I can stack the three properties together to solve for the form of the Z Transform (similar to how I apply multiple properties of the Laplace and Fourier Transform Simultaneously)