1. **Problem statement:** Find the residue of the function $$f(z) = \frac{z e}{(z-1)^3}$$ at its poles. 2. **Identify the poles:** The function has a pole at $$z=1$$ of order 3 because of the denominator $$(z-1)^3$$. 3. **Residue formula for a pole of order $m$:** For a pole of order $m$ at $z=a$, the residue is given by $$\text{Res}(f,a) = \frac{1}{(m-1)!} \lim_{z \to a} \frac{d^{m-1}}{dz^{m-1}} \left[(z-a)^m f(z)\right].$$ 4. **Apply the formula:** Here, $m=3$ and $a=1$, so $$\text{Res}(f,1) = \frac{1}{2!} \lim_{z \to 1} \frac{d^2}{dz^2} \left[(z-1)^3 \frac{z e}{(z-1)^3}\right] = \frac{1}{2} \lim_{z \to 1} \frac{d^2}{dz^2} (z e).$$ 5. **Simplify inside the derivative:** Since $e$ is a constant, $$\frac{d^2}{dz^2} (z e) = e \frac{d^2}{dz^2} z = e \cdot 0 = 0.$$ 6. **Conclusion:** The residue at the pole $z=1$ is $$\boxed{0}.$$