Why amplitudes are rational functions? In Bootstrap and Amplitudes: A Hike in the Landscape of Quantum Field Theory there are few statements about analytical structure of amplitudes.

I want to understand statement:

Tree amplitudes must be rational functions of Mandelstam
variables.

What is the reason for such a property of tree level amplitudes? What is the simplest explanation of such a fact?
Comments about other analytical properties are also very appreciated!
 A: After the OP explained in the comments what exactly they're looking for, I will attempt an answer. There are a few separate facts that need explanation:

*

*Tree amplitudes are rational functions of Mandelstam invariants: This is an obvious statement, that I have explained in the comments, and I will repeat here to preserve: The tree level feynman rules contain two elements: the vertex factors and propagators. Locality implies that vertex factors are always polynomials in momenta (since we are allowed to keep only a finite number of derivates in the action). For the same reason, propagators are also polynomials in momenta. This implies that amplitudes(vertex factor/propogators) are rational functions in mandelstam variables.


*The poles are simple poles in Mandelstam invariants provided we have a massless, local theory: This requires a little explanation, so I will explicitly write down the line of reasoning that leads to this conclusion.
If we have a local, massless theory, the denominator is always quadratic in momenta. This is a totally nontrivial point since naively, we could think of interactions of the form $\mathcal{L}_{int} \sim g_n \phi \partial ^n \phi$. But of course, massless theories do not admit such terms bilinear in fields since generic values of the $\{g_i\}$ introduces new poles in the two point function, that is to be interpreted as a new massive particle. Therefore all propagators are of the form $\frac{i}{(k_{i_1}+k_{i_2}+k_{i_3}...+k_{i_m})^2}$. No two propagators of a tree diagram carry the same momenta(for generic values of the external momenta), and hence when this propagator goes on shell, we find a simple pole in the Mandelstam variable $s_{{i_1 i_2....i_m}}=(k_{i_1}+k_{i_2}+k_{i_3}...+k_{i_m})^2$.
A: @Anonjohn has given a good answer. I would like to add a bit more. I think, that the author of statement tacitly assumes dealing with massless theory with a linear dispersion relation (kinetic term of form $\phi \Box \phi $) : Why does nature favour the Laplacian?. So each propagator is of the form:
$$
\frac{1}{(\sum_{i \in I} k_i)^2}
$$
The analytical properties are crucial in determining the exact expression for tree amplitudes in massless theories. For instance, the proof of famous Parke-Taylor formula by BCFW https://arxiv.org/pdf/hep-th/0501052.pdf - is based on them.
For a good introduction I recommend these notes, written by the same author as the article you cite - https://arxiv.org/pdf/1308.1697.pdf.
