Laurent series centered at $z_0=1$ with $1<|z-1|<\infty$ There is a function $f(z)=\dfrac{1}{z(z-1)}$, and it can be expanded as $\dfrac{1}{z-1}=\dfrac{1}{z}\cdot \dfrac{1}{1-\frac{1}{z}}$. Furthermore, $\dfrac{1}{1-\frac{1}{z}}=1+\dfrac{1}{z}+\dfrac{1}{z^2}...$ I substituted $w=z-1$ and expanded to get my Laurent series: $\dfrac{1}{(z-1)^2}+\dfrac{1}{(z-1)^3}+\cdots$. Did I approach this correctly?
 A: *

*When the function is rational, the standard procedure is to start with finding its expansion into partial fractions. In the present case, you get $$f(z)=\dfrac{1}{z(z-1)}=\dfrac{1}{z-1}-\dfrac{1}{z}.$$ Hence, making use of your substitution $w=z-1$ (or $z=1+w$), we obtain
\begin{align}
\dfrac{1}{z-1}-\dfrac{1}{z}&=\frac{1}{w}-\frac{1}{w+1}=\frac{1}{w}-\frac{1}{w}\frac{1}{1-(-\dfrac{1}{w} ) }\\&=\frac{1}{z-1}-
\frac{1}{z-1}\frac{1}{1-(-\dfrac{1}{z-1})}.\tag 1
\end{align}


*And so, you just need to expand $\dfrac{1}{1-( -\dfrac{1}{w}
) }=\dfrac{1}{1-( -\dfrac{1}{z-1}) }$
for $|w|=|z-1|>1$:
$$\frac{1}{1-(-\dfrac{1}{z-1})}=\sum_{n\geq 0}\frac{(-1)^{n}}{(z-1)^{n}}=1+\sum_{n\geq 1}\frac{(-1)^{n}}{(z-1)^{n}}. \tag 2 $$
Please notice that this is an alternating series.


*Therefore, combining $(1)$ and $(2)$, for $|z-1|>1$ the original function may be expanded into
\begin{align}
f(z)&=\frac{1}{z-1}-\frac{1}{z-1}-\frac{1}{z-1}\sum_{n\geq 1}\frac{(-1)^{n}}{\left( z-1\right) ^{n}} \tag 3\\&=\sum_{n\geq 1}\frac{(-1)^{n+1}}{\left( z-1\right) ^{n+1}}=\sum_{n\geq 2}\frac{(-1)^{n}}{\left( z-1\right) ^{n}}\\&=\frac{1}{(z-1)^2}-\frac{1}{(z-1)^3}\pm\cdots\tag{4}
\end{align}
A: It is not clear for what you are using $w = z-1$  and the link between $f(z)$ and the remainder of your argument is unclear.  However, you have nearly got the right answer.
Instead, you could justify the result by writing,
\begin{align}
f(z) &= \frac{1}{z(z-1)} \\
&= \frac{1}{(z-1)^2(1+\frac{1}{z-1})}
\end{align}
and when $\lvert z-1 \rvert > 1$ the second parentheses in the denominator can be expanded as a convergent infinite series, so that,
\begin{align}
f(z) &= \frac{1}{(z-1)^2} \cdot \Big(1 - \frac{1}{z-1}+\frac{1}{(z-1)^2}-\cdots\Big) \\
&=\frac{1}{(z-1)^2} - \frac{1}{(z-1)^3}+\frac{1}{(z-1)^4} -\cdots
\end{align}
which is nearly the same as your result, derived directly from the original expression. Notice the alternating sign in the sum.
