Recently when I read one paper "DISSECTING A POLYGON INTO TRIANGLES" by Catalan, it uses the General Binomial Theorem in the expansion of $(1+x)^{\frac{1}{2}}$.
Then I refer to proofwiki proof 1, and I have question about how the 2nd equality got when $f(x)=(1+x)^{\alpha}=\sum\limits_{n=0}^{\infty}\frac{\prod\limits_{k=0}^{n-1}(\alpha-k)}{n!}x^n$ and $D_x f(x)=\sum\limits_{n=0}^{\infty}\frac{\prod\limits_{k=0}^{n-1}(\alpha-k)}{n!}n x^{n-1}$: $$ \begin{align*} (1+x)D_x f(x)&=\sum\limits_{n=1}^{\infty}\frac{\prod\limits_{k=0}^{n-1}(\alpha-k)}{(n-1)!}x^{n-1} +\sum\limits_{n=1}^{\infty}\frac{\prod\limits_{k=0}^{n-1}(\alpha-k)}{(n-1)!}x^{n}\\ &=\alpha+\sum\limits_{n=1}^{\infty}(\frac{\prod\limits_{k=0}^{n}(\alpha-k)}{n!}+\frac{\prod\limits_{k=0}^{n-1}(\alpha-k)}{(n-1)!})x^{n} \end{align*} $$ To get the above 2nd equality, it is to prove $$ \alpha=\sum\limits_{n=1}^{\infty}\frac{\prod\limits_{k=0}^{n-1}(\alpha-k)}{(n-1)!}x^{n-1}(1-\frac{(\alpha-n)x}{n}) $$
But how to proceed then?
Q:
How to get the above 2nd equality $\alpha+\sum\limits_{n=1}^{\infty}(\frac{\prod\limits_{k=0}^{n}(\alpha-k)}{n!}+\frac{\prod\limits_{k=0}^{n-1}(\alpha-k)}{(n-1)!})x^{n}$?
$\begin{align}\sum\limits_{n=1}^{\infty}\frac{\prod\limits_{k=0}^{n-1}(\alpha-k)}{(n-1)!}x^{n-1}=&\alpha+\sum\limits_{n=2}^{\infty}\frac{\prod\limits_{k=0}^{n-1}(\alpha-k)}{(n-1)!}x^{n-1}\\ =&\alpha+\sum\limits_{m=1}^{\infty}\frac{\prod\limits_{k=0}^{n-1}(\alpha-k)}{m!}x^{m}\end{align}$
The first equality is by expanding the LHS and the second equality is by changing $n-1$ to $m$.