# 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. The screenshot shows their working, which I believe restricts to tree-level amplitudes only. Is there any contradiction?
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.
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.