I was learning Laurent development (Conway's text, page 107), where Conway uses a two-page (a little tedious) proof to show the existence of Laurent Series, and uniqueness is left as exercise. Then I checked Ahlfors but he uses the same procedure. But I think the proof can be greatly simplified as follows:
First, for any $z\in\text {ann}(a;R_1,R_2)$, choose $r_1,r_2$ s.t. $R_1<r_1<|z-a|<r_2<R_2$, and define $\gamma_1=r_1 e^{-i\theta},\gamma_2=r_2 e^{i\theta},\theta\in[0,2\pi]$. We have $$f(z)=\frac{1}{2\pi i}\int_{\gamma_1}\frac{f(w)}{w-z}\text d w+\int_{\gamma_2}\frac{f(w)}{w-z}\text d w$$ by Cauchy's Integral Formula. Now, we can expand the integrand using geometric series, then interchange the sum and integral since the series is uniformly convergence on $\{\gamma_1\},\{\gamma_2\}$: \begin{align}f(z)=&\frac{1}{2\pi i}\int_{\gamma_1}\frac{f(w)}{z-a}\sum_{k\geq 0} (\frac{w-a}{z-a})^k\text d w+\frac{1}{2\pi i}\int_{\gamma_2}\frac{f(w)}{w-a}\sum_{k\geq 0}(\frac{z-a}{w-a})^k\text d w\\=&\sum_{k\leq-1}[\frac{1}{2\pi i}\int_{-\gamma_1}\frac{f(w)}{(w-a)^{k+1}}\text d w] (z-a)^k+\sum_{k\geq 0}[\frac{1}{2\pi i}\int_{\gamma_2}\frac{f(w)}{(w-a)^{k+1}}\text d w] (z-a)^k \end{align} But the integral $a_k=\frac1{2\pi i}\int_{\gamma}\frac{f(w)}{(w-a)^{k+1}}\text d w$ is independence of the choice of $\gamma=re^{i\theta}$, hence we obtained the desired expansion.
To prove the uniqueness, suppose $f(z)=\sum_k a_k(z-a)^k$, then the convergence must be uniform and absolute, hence we can integrate each item of $f/(z-a)^{n+1}$ to identify $a_n$: $$\frac1{2\pi i}\int_\gamma \frac{f(z)}{(z-a)^{n+1}}\mathrm dz=\frac1{2\pi i}\int_\gamma\frac{a_n}{z-a}\mathrm dz=a_n$$ Done.
This proof uses the same method as Conway's proof of power series expansion of analytic functions, and is really straightforward and intuitive. I don't know if it's correct. And why both Conway and Ahlfors use a more abstract and tedious way to develop the Laurent Series? Feel free to point out any errors or share your understanding on this.