Prove that $z^{\alpha} = \sum_{k=0}^{\infty} {\alpha \choose k} z_{0} ^{ \alpha - k}(z -z_{0})^k$ $$z^{\alpha} = \sum_{k=0}^{\infty} {\alpha \choose k} z_{0}  ^{ \alpha - k}(z -z_{0})^k$$
for $z,z_{0} > 0$ with $ |z - z_{0}| < |z_{0}|$
I assume that I should use differentiation
$\alpha \cdot z^{\alpha-1} = $ but I'm not sure that this is the right for to the solution and how does the RHS look after differentiating it. I'm thankful for every hint.
 A: It is sufficient to prove that $$z^\alpha=\sum_{k\ge0}{\alpha \choose k}(z-1)^k.\tag1$$
This is because
$$\begin{align}
z^\alpha&=z_0^\alpha\left(\frac{z}{z_0}\right)^\alpha \\
&=z_0^\alpha\sum_{k\ge0}{\alpha\choose k}\left(\frac{z}{z_0}-1\right)^k\\
&=\sum_{k\ge0}{\alpha\choose k}z_0^{\alpha-k}\left(z-z_0\right)^k.
\end{align}$$
Here is a proof for $(1)$. Let $$f(z)=(z+1)^\alpha.$$
We will write
$$f(z)=\sum_{k\ge0}\frac{f^{(k)}(0)}{k!}z^k.$$
Let $D$ be differentiation w.r.t. $z$, $D:\phi(z)\mapsto\phi'(z)$.
Then we have
$$\begin{align}
D^0f&=(z+1)^\alpha\\
D^1f&=\alpha (z+1)^{\alpha-1}\\
D^2f&=\alpha(\alpha-1)(z+1)^{\alpha-2}\\
D^3f&=\alpha(\alpha-1)(\alpha-2)(z+1)^{\alpha-3}\\
&...\\
D^kf&=p_k(\alpha)(z+1)^{\alpha-k},
\end{align}$$
where $$p_k(\alpha)=\prod_{j=0}^{k-1}(\alpha-j),$$
and $p_0(\alpha)=1$. Thus $f^{(k)}(0)=D^kf|_{z=0}=p_k(\alpha)$ and
$$f(z)=\sum_{k\ge0}\frac{p_k(\alpha)}{k!}z^k.$$
Recall the definition $${n\choose k}=\prod_{i=1}^{k}\frac{n-i+1}{i}=\frac{\prod_{i=1}^{k}(n-i+1)}{\prod_{i=1}^{k}i}=\frac{\prod_{i=0}^{k-1}(n-i)}{k!}=\frac{p_k(n)}{k!}.$$
Thus
$$\frac{p_k(\alpha)}{k!}={\alpha\choose k},$$
and $$(1+z)^\alpha=\sum_{k\ge0}{\alpha\choose k}z^k.$$
Replace $z$ with $z-1$, and get
$$z^\alpha=\sum_{k\ge0}{\alpha\choose k}(z-1)^k.$$
We are done.
A: We obtain for $z_0\ne 0$ according to the binomial series expansion:
\begin{align*}
\color{blue}{z^{\alpha}}&=\left(z_0+(z-z_0)\right)^\alpha\\
&=z_0^{\alpha}\left(1+\frac{z-z_0}{z_0}\right)^{\alpha}\\
&=z_0^{\alpha}\sum_{k=0}^{\infty}\binom{\alpha}{k}\left(\frac{z-z_0}{z_0}\right)^k\\
&\,\,\color{blue}{=\sum_{k=0}^{\infty}\binom{\alpha}{k}\left(z-z_0\right)^{k}z_0^{\alpha-k}\qquad\qquad\left|\frac{z-z_0}{z_0}\right|<1}
\end{align*}
