Mass dimension of an $n$-particle scattering amplitude in 4D For the 4-dimensional case, and using the cross-section formula, how can we show that the mass dimensions of an $n$-particle amplitude must be $$[A_n] = 4-n~?\tag{2.99}$$ 
My understanding is that the cross-section must have dimensions of an area, but I don't quite understand how I can then find the dimensions of a scattering amplitude. 
I am assuming that the differential cross-section is the amplitude squared, and trying to work backwards.
$$\frac{d\sigma}{d\Omega}=|A|^2,$$ where $A$ is the amplitude. Is this the correct path? 
Reference: https://arxiv.org/abs/1308.1697, equation 2.99.
EDIT: Allow me to add a screenshot which may clarify some my confusion. After reading the answer below and being comfortable with the methods (and finding some agreeing literature in a PhD thesis and other textbooks), I am still not sure why this doesn't match up to the paper I'm using. There's a lot of funky conventions, which I suppose is a hazard of dealing with amplitudes.
Please see this screenshot:
 The screenshot shows their working, which I believe restricts to tree-level amplitudes only. Is there any contradiction?
 A: The claimed result $[A_n]=4-n$ is correct, and so is the reasoning of Helvang and Huang in the quoted text in the OP. Notice that $n$ is the total number of particles involved in the process, the in+out particles. 
In particular, the mass dimension certainly does not depend on the loop order that is needed to generate $A_n$, and so one is free to determine $[A_n]$ by reasoning with tree-level amplitudes only, and Huang and Elvang do so. 
But let me present an alternate derivation, just for the sake of discussion.
The mass dimension $[A_n]$ is as well independent of the spin of the particles involved in the process, and so I can calculate $[A_n]$ for any spin just by calculating $[A_n]$ for processes involving spin-0 only. Moreover, it's independent for particle or antiparticles, so I will use just a real scalar field. 
I do so using the LSZ prescription: Fourier transform the correlator $\langle \Phi(x_1)\ldots \Phi(x_n)\rangle$, look at the $n$ one-particle simple poles by multipling it for $\prod_{i=1}^n p_i^2-m_i^2$, and remove a $\delta^4(\sum p)$ for passing from $S$ to $A_n$, namely $S=1+(2\pi)^4i\delta^4(\sum p)A_n$. 
Since each $\Phi$ adds $[\Phi(x)]=1$, each Fourier transform adds $[\int d^4 x]=-4$, each residue adds $[p_i^2-m_i^2]=2$, and removing the delta function adds $[1/\delta^4(\sum p)]=4$, we get 
$$
[A_n]=n-4n+2n+4=4-n
$$
as claimed. 
