<s>[INST] 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?
 [/INST] 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.
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