Uniqueness of series representation. (Complex Analysis). Theorem : If a series $$ \sum_{n=0}^\infty a_n (z-z_0)^{n}$$
converges to $f(z)$ at all points interior to some circle $|z-z_0| = R$ , then it is the taylor series expansion for $f$ in powers of $z-z_0$.
(1) First we write the series by changing the index as : $$\sum_{m=0}^\infty a_m (z-z_0)^{m}$$
(2) Now we make use of another theorem ( I don't want to get into that , just accepting its result and writing the following) : $$\int_{C}g(z)f(z)dz = \sum_{m=0}^\infty a_m \int_{C}g(z)(z-z_0)^{m}dz$$
where $g(z) = \dfrac{1}{2 \pi i}\cdot \dfrac{1}{(z-z_0)^{n+1}}$ and $C$ is some circle centered at $z_0$ and with radius less than $R$.
(3) Next we prove that : $$\int_{C}g(z)f(z)dz = \dfrac{1}{2 \pi i}\cdot \int_{C}\dfrac{f(z)dz}{(z-z_0)^{n+1}} = \dfrac{ f^{(n)}(z_0)}{n!}$$ 
I was able to prove it using the extension of Cauchy Integral formula. So we move to next step.
(4) Now we prove that : $$\int_{C}g(z)(z-z_0)^{m}dz = \dfrac{1}{2 \pi i}\cdot \int_{C}\dfrac{dz}{(z-z_0)^{n-m+1}} =  \begin{cases} 
      0 ,\: \: when \:\: m \neq n \\
      1 ,\: \: when \:\: m=n \\ 
\end{cases} $$
I was able to prove this too.. Now my problem arises in the next step.. 
(5) Now this step says , " it is clear that " : $$\sum_{m=0}^\infty a_m \int_{C}g(z)(z-z_0)^{m}dz = a_n$$ , 
How can we write this ? 
If we take $\int_{C}g(z)(z-z_0)^{m}dz = 1$ (say) , then ,
$\dfrac{ f^{(n)}(z_0)}{n!} = \sum_{m=0}^{\infty}a_m \cdot 1$ ,
How can we write step (5) , could anyone help ?
P.S : I know its a long proof , but please , bear with me.. !!

From (2) you have
${1 \over 2 \pi i} \int_C {1 \over (z-z_0)^{n+1}} f(z)dz = {1 \over 2 \pi i}\sum_m a_m \int_C {(z-z_0)^m \over (z-z_0)^{n+1}}dz $, and with (4) you have shown that
${1 \over 2 \pi i} \int_C {(z-z_0)^m \over (z-z_0)^{n+1}}dz = \delta_{mn}$, where $\delta_{mn} = \begin{cases} 1 , & m=n\\
0, & \text{otherwise} \end{cases}$. So, now substitute this result into the first
to get
${1 \over 2 \pi i}\sum_m a_m \int_C {(z-z_0)^m \over (z-z_0)^{n+1}}dz = \sum_m a_m \delta_{mn} = a_n$.
The combining this with the result from (3) gives
$\int_C g(z)f(z)dz = a_n = {1 \over n!} f^{(n)}(z_0)$.
