 Solution

If the pole has order $k+1$, multiplying the Laurent series by $(z-a{)}^{k+1}$ gives a function $h$ with a holomorphic extension across $a$:Its coefficient of $(z-a{)}^k$ is exactly $c_{-1}$. The Taylor series coefficient formula gives $h^{(k)}(a)=k!c_{-1}$. Continuity of this derivative at $a$ therefore proves