how can I prove the following for positive and negative integers as power $a$ as well as rational powers $a$? I would prefer to prove it without differentiation and Taylor's theorem if that's possible. I think for positive integers it might be possible to solve it with induction but I can't come to any solution.
$$ (1+x)^a = \sum_{k=0}^\infty P_k(a)x^k $$
$ P_k(a) = \frac1{k!}a(a-1)...(a-k+1) $
($P_0(a)=1$)
Let’s use the Stirling numbers oft the first kind $\,\displaystyle {n\brack m}$
(e.g. https://en.wikipedia.org/wiki/Stirling_numbers_of_the_first_kind).
We have $\enspace\displaystyle \frac{(-1)^v}{v!} \sum\limits_{n=0}^v {v\brack n} (-a)^n = \frac{(-1)^v}{v!} (-a)(-a+1)…(-a+v-1) = {\binom a v}$
and because of $\,\displaystyle {v\brack n}=0\,$ for $\,n>v\,$ it’s $\enspace\displaystyle \frac{(-1)^v}{v!} \sum\limits_{n=0}^\infty {v\brack n} (-a)^n = {\binom a v} \,$ .
With $\,|x|<1\,$ follows:
$\enspace\displaystyle \sum\limits_{v=0}^\infty {\binom a v} x^v = \sum\limits_{v=0}^\infty \frac{(-1)^v x^v}{v!} \sum\limits_{n=0}^\infty {v\brack n} (-a)^n = \sum\limits_{n=0}^\infty a^n \sum\limits_{v=0}^\infty (-1)^{n-v}{v\brack n}\frac{x^v}{v!} $
$\hspace{2.2cm}\enspace\displaystyle = \sum\limits_{n=0}^\infty \frac{a^n}{n!} (\ln(1+x))^n = e^{a\ln(1+x)}=(1+x)^a $