Q:

Raise a power series to a fractional exponent?

In showing that $\log^\alpha{(1+x)}$ is $O((x)^\alpha)$ at $1$, for $\alpha>0$, one can note that 
$$\left ( \frac{\log{(1+x)}}{x} \right )^\alpha \overset{x\to 0}{\longrightarrow} \left ( 1\right )^\alpha = 1.$$
So we know that $$\log^\alpha{(1+x)}= (x)^\alpha + o((x)^\alpha).$$
But how would I find $\beta > \alpha$ and $c\in \mathbb{R}$ such that  
$$\log^\alpha{(1+x)}= (x)^\alpha + c(x)^\beta + o((x)^\beta)?$$
I'd like to raise the power series to an exponent:
$$\log^\alpha(1+x) = \left ( x - x^2/2 + x^3/3 + \dotsb \right) ^ \alpha = \,\,\,??$$
Is there a version of the multinomial theorem I need to use here? This page I found gives me a formula for a multinomial when the first term is larger than the sums of the rest of the terms, but I don't think I want to use that...

A:

Landscape has provided a very nice answer in the comments. I will summarize in this community wiki, hoping I am getting it right:
Given a power series to a (possibly fractional) exponent $\alpha > 0$, say the one in the original question:  
$$ \left ( x - x^2/2 + x^3/3 + \dotsb \right) ^ \alpha,$$
one can factor out the $x$ term, and truncate at whatever term one desires
$$ x^\alpha \left  ( 1 - x/2 + x^2/3 + o(x^2) \right) ^ \alpha$$
and then apply the generalized binomial theorem
$$= x^\alpha \left  ( 1 + \alpha\cdot[- x/2 + x^2/3 + o(x^2) ] + \frac {\alpha (\alpha - 1) } {2} [ - x/2 + x^2/3 + o(x^2)  ]^2 \\
+ o( [ - x/2 + x^2/3 + o(x^2) ]^2)  \right)  \\
= x^\alpha - \frac {\alpha} {2} x^{\alpha + 1} + \left ( \frac {\alpha} {3} + \frac {\alpha(\alpha-1)} {8}  \right )x^{\alpha + 2} + o(x^{\alpha  + 2}).$$

