In an example in our notes it says: Compute the Laurent series for $f(z)= \frac{1}{z^2(z+1)}$ and determine the annulus of convergence.
No more information was provided. So I did it on my own by factoring it into $\left( \frac{1}{z^2} \right) \left( \frac{1}{z(1-(-(1/z)))} \right)$ then using geometric series to get an expression where it's convergent for $|z| > 1$. I looked back at the notes to check my answer and saw the way he obtained the solution was by factoring it into $\left( \frac{1}{z^2} \right)(1-(-z))$ and getting an expression where it was convergent for all $|z| < 1$. Would they both be considered correct answers?
Your Laurent series would have two different representations in two different sectors: |z| < 1 and |z| > 1. Since you make the decomposition into series around zero, and you have a pole in zero, you will always have negative terms. If you took, for exemple, $$\frac{1}{(z - 1)(z - 2)},$$ you would have 3 different decompositions: one for |z| < 1, which is just a Tailor series (with n=0 to infinity), one for $1 < |z| < 2$ (both negative and positive degrees) and one for $|z| > 2$ (as far as I remember, this one contains only the negative degrees of z).
In your case, the Taylor series "shrink" to an empty circle with radius 0 because of the pole in zero.
So, both of your solutions are correct, but for different regions.
BTW, when you say $|z| < 1$, you should state that $z\neq 0$: $0< |z| < 1$