= 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 $a$:
$$
h(z)=c_{-(k+1)}+c_{-k}(z-a)+\cdots+c_{-1}(z-a)^k+\cdots.
$$
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
$$
\boxed{\operatorname{res}(f,a)=\frac{h^{(k)}(a)}{k!}=\lim_{z\to a}\frac{h^{(k)}(z)}{k!}}.
$$
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