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...