The theorem and the first part of its proof is shown below:
In particular, the authors conclude (2 lines below equation (2)) that
$(i): P_R(n) = {n + a_1 \choose a_1}+\cdots+ {n+a_r -(r-1) \choose a_r} + c$.
Question 1: Is there an immediate way to see how they arrived at this equation?
Question 2: It is not clear at first sight that $c$ is even an integer. Or it is? Anyway, the authors consider it obvious that $c$ is an integer and they proceed to show that it is non-negative via a rather complicated argument.
In trying to understand equation $(i)$ i came up with the following argument, which not only shows that $c$ is an integer, but that it is also non-negative:
My argument: There exists a unique expression
$(ii): P_R(n) = {n + a_1 \choose a_1}+{n + a_2-1 \choose a_2}+\cdots+ {n+a_s -(s-1) \choose a_s}$
with $a_1 \ge a_2 \ge \cdots \ge a_s \ge 0$. As in the argument in the proof, suppose that $P_R(n) - P_R(n-1)$ has a unique expression
$P_R(n) - P_R(n-1) = {n + b_1 \choose b_1}+\cdots+ {n+b_r -(r-1) \choose b_r}$
with $b_1 \ge b_2 \ge \cdots \ge b_r \ge 0$. Then using Pascal's triangle we must have that
${n + b_1 \choose b_1}+\cdots+ {n+b_r -(r-1) \choose b_r} = {n + a_1-1 \choose a_1-1}+\cdots+ {n+a_s-1 -(s-1) \choose a_s-1}$.
By the uniqueness of the expression on the left and the fact that $b_i \ge 0$ we can identify $a_i = b_i + 1$ for $i=1,\dots,r$ and we must also have that
${n + a_{r+1}-1-r \choose a_{r+1}-1}+\cdots+ {n+a_s-1 -(s-1) \choose a_s-1}=0$.
The only way this last equation can be satisfied is if $a_{r+1}=\cdots=a_s=0$. Substituting this in $(ii)$ we get precisely equation $(i)$, where $c:=s-r \ge 0$.
Question 3: Any comments regarding my argument?
It seems they used $P_R(k) - P_R(k-1) = {k + b_1 \choose b_1}+\cdots+ {k+b_r -(r-1) \choose b_r}$ by summing these up for $k=1,\dots,n$ and using the well known combinatorial identity $\sum_{k=0}^n{k + b_1 \choose b_1}={n + b_1 + 1\choose b_1 + 1}$. Then $c$ is nothing but $P_R(0)-1$ which is obviously an integer.