Power series of a function related to Gamma function
Using the binomial series expansion and applying the binomial identity $$\binom{-\alpha}{n}=\binom{\alpha+n-1}{n}(-1)^n$$
we derive for $|z|<1$ and $\alpha\in\mathbb{C}$ \begin{align*} f(z)=\frac{1}{(1-z)^\alpha}&=\sum_{n=0}^\infty\binom{-\alpha}{n}(-z)^n\\ &=\sum_{n=0}^\infty\binom{\alpha+n-1}{n}z^n\\ \end{align*}
It follows \begin{align*} a_n(\alpha)&=\binom{\alpha+n-1}{n}=\frac{1}{n!}(\alpha+n-1)(\alpha+n-2)\cdots\alpha\\ &=\frac{1}{n!}(\alpha+n-1)^{\underline{n}} \end{align*} with $z^{\underline{n}}=z(z-1)\cdots(z-n+1)$ the falling factorial.
The following is valid \begin{align*} \lim_{n\to\infty}\frac{a_n(\alpha)}{n^{\alpha-1}}=\lim_{n\to\infty}\frac{(\alpha+n-1)^{\underline{n}}}{n!n^{\alpha-1}}=\frac{1}{\Gamma(\alpha)}\tag{1} \end{align*}
In order to show (1) we use a representation of the Gamma function $\Gamma(\alpha)$ according to C.F. Gauss. We obtain \begin{align*} \color{blue}{\Gamma(\alpha)}&=\lim_{n\to\infty}\frac{n^\alpha n!}{\alpha(\alpha+1)\cdots(\alpha+n)}\\ &=\lim_{n\to\infty}\frac{n^\alpha n!}{(\alpha+n)^{\underline{n+1}}}\\ &=\lim_{n\to\infty}\left(\frac{n}{\alpha+n}\cdot\frac{n^{\alpha-1} n!}{(\alpha+n-1)^{\underline{n}}}\right)\\ &=\lim_{n\to\infty}\frac{1}{\frac{\alpha}{n}+1}\cdot\lim_{n\to\infty}\frac{n^{\alpha-1} n!}{(\alpha+n-1)^{\underline{n}}}\\ &\color{blue}{=\lim_{n\to\infty}\frac{n^{\alpha-1} n!}{(\alpha+n-1)^{\underline{n}}}}\\ \end{align*} and the claim follows.