Showing $|x|^\alpha$ is an $A_p$ weight for $-d<\alphaWe say $w:\mathbb{R}^d \to [0,\infty)$ is an $A_p$ weight if
$$\frac{1}{|B|}\int_B w \left(\frac{1}{|B|}\int_Bw^{1-p'}\right)^{p-1} \le C(p)$$ for all balls $B$, and where $p'$ is the holder conjugate of $p>1$. We want to show $|x|^\alpha$ is an $A_p$ weight if and only if $d<\alpha<d(p-1)$. Here is my progress:
First take balls centered at the origin. We have
$$\frac{1}{|B(0,r)|}\int_{B(0,r)}w(x) \ dx =\frac{1}{c(d)r^d} \int_0^r \int_{\partial B(x,s)} |x|^\alpha \ dS(x) \ ds =\frac{1}{c(d)r^d} \int_0^r dc(d) s^{\alpha + d-1} \ ds$$
converges if and only if $\alpha >-d$ (i.e., $r^c$ is integrable at $0$ if and only if $c>-1$), and in this case is equal to 
$$= \frac{d}{\alpha + d}r^{\alpha} <\infty$$
And then
$$\left(\frac{1}{|B(0,r)|}\int_{B(0,r)} w^{1-p'} \right)^{p-1} = \left(\frac{1}{c(d)r^d} \int_0^r \int_{\partial B(0,s)} s^{\alpha - \alpha p'} \ dS(x) \ ds\right)^{p-1}$$
$$= \left(\frac{1}{c(d)r^d} \int_0^r dc(d) s^{\alpha - \alpha p' + d-1} \ ds\right)^{p-1}$$
$$ =\left(\frac{d}{\alpha - \alpha p ' + d}\right)^{p-1} r^{\alpha (1-p')(p-1)}$$
$$=\left(\frac{d}{\alpha - \alpha p ' + d}\right)^{p-1} r^{-\alpha}$$
where the last integral converges if and only if $\alpha(1-p') + d = \alpha \cdot \displaystyle \frac{-1}{p-1} +d >0$, or, if $\alpha <d(p-1)$. Then we see multiplying the two expressions we get a constant.
So, we definitely have necessity of our range. Sufficiency is a bit harder, because we have to show it for any balls.
For balls $B(y,r)$ with $|y|\le r$ we can simply extend the domain of integration to get
$$\frac{1}{c(d)r^d}\int_{B(y,r)}\left(\frac{1}{c(d)r^d}\int_{B(0,r)}|x|^{\alpha(1-p')}\right)^{p-1}$$
$$\le \frac{C}{r^{dp}} \int_{B(0,|y|+r)}|x|^\alpha \left(\int_{B(0,|y|+r)}|x|^{\alpha(1-p')}\right)^{p-1}$$
$$\le C\frac{(|y|+r)^{dp}}{r^{dp}} \le C$$
since we use $|y| \le r$.
But I don't know how to extend to balls with $|y|>r$. Any tips? In general I just don't know how to deal with these integrals over balls not centered at origin.

Actually you have taken care of the most "troublesome" balls already!
Fist observe that your argument for balls $|y|\le r$ works also if $|y|\le Cr$ for some constant $C>1$ (imagine $C=10^{10}$). 
That means we may assume $|y|>Cr$. That is, the radius of the ball is small (imagine tiny) compared to its distance from the origin. 
As a consequence, we expect the average $\frac1{|B|}\int_B |x|^\beta$ to be roughly $|y|^\beta$ for all real $\beta$. 
Making this precise, I'm claiming that we have $\frac1{|B|}\int_B |x|^\beta\lesssim |y|^\beta$ (try to prove this). 
(Here, $\lesssim$ denotes $\le$ up to a multiplicative constant.)
This implies
$$\left(\frac1{|B|}\int_B w\right)\left(\frac{1}{|B|}\int_B w^{1-p'}\right)^{p-1} \lesssim |y|^\alpha |y|^{\alpha (1-p')(p-1)} = 1.$$
We didn't even need to make use of the constraint on $\alpha$ in this case.
