Constant term extraction using combinatorial Nullstellensatz $\DeclareMathOperator\CT{CT}$Given a Laurent polynomial $g$, let $\CT(g)$ denote its constant term.
Consider the specific Laurent polynomial
$$f_n(x_1,\dots,x_r)=\left(1+\prod_{j=1}^r(1+x_j)+\prod_{j=1}^r\left(1+\frac1{x_j}\right)\right)^n.$$

QUESTION. Is there a Combinatorial Nullstellensatz (of Noga Alon) proof of this?
$$\CT(f_n)=\sum_{m=0}^n\binom{n}m\sum_{k=0}^m\binom{m}k^{r+1}.$$

 A: The argument below works only for odd $r$. This is annoying, but I do not see how to modify it for even $r$. Of course, there is a simple argument without CN, but I guess it is not what you ask for.
We start with denoting $x_i=y_{i}/y_{i-1}$ for $i=1,\ldots,r$, where $y_0$, $\ldots$, $y_r$ are new variables. It is easy to see that the constant term with respect to $x_i$'s is the same as with respect to $y_j$'s (in other words, any non-constant monomial in $x_i$'s is non-constant in $y_i$'s too). Our Laurent polynomial reads as $(1+Z)^n$, where $$Z=\prod_{i=1}^r\left(1+\frac{y_i}{y_{i-1}}\right)+\prod_{i=1}^r\left(1+\frac{y_{i-1}}{y_i}\right)=
\frac{(y_0+y_1)(y_1+y_2)\ldots (y_{r-1}+y_r)(y_r+y_0)}{y_0\ldots y_r},$$
and we get $${\rm CT}\left( (1+Z)^n\right)=\sum_{m=0}^n {n\choose m} {\rm CT} (Z^m)=\sum_{m=0}^n {n\choose m} \left[y_0^m\ldots y_r^m\right] (y_0+y_1)^m\ldots (y_r+y_0)^m,$$
and for computing these coefficients we may try to apply Combinatorial Nullstellensatz in the explicit form. (Here we could simply say that if we take the monomial $y_0^ky_1^{m-k}$ from $(y_0+y_1)^m$, we must take $y_i^ky_{i+1}^{m-k}$ from each $(y_i+y_{i+1})^m$, thus the result. But our goal is to apply CN, so we do not do this$^{*}$.)
Assume that $r$ is odd. Then we replace each binomial $(y_{2i}+y_{2i+1})^m$ to $\prod_{j=0}^{m-1} (y_{2i}+y_{2i+1}-j)$ and each binomial $(y_{2i-1}+y_{2i})^m$ to $\prod_{j=0}^{m-1} (y_{2i-1}+y_{2i}-m-1-j)$ (here $y_{r+1}:=y_0$ as usually). Denote the product of new deformed binomials by $G_m(y_0,\ldots,y_r)$ and look at the values of $G_m(a_0,\ldots,a_r)$ where $a_i\in \{0,1,\ldots,m\}=:A$. If $G_m(a_0,\ldots,a_r)\ne 0$, we must have $a_0+a_1\geqslant m$, $a_1+a_2\leqslant m$, $a_2+a_3\geqslant m$, $\dots$, $a_r+a_0\leqslant m$. Then $$m(r+1)/2\leqslant (a_0+a_1)+(a_2+a_3)+\ldots+(a_{r-1}+a_r)=\sum a_i\\
=(a_1+a_2)+(a_3+a_4)+\ldots+(a_r+a_0)\leqslant m(r+1)/2,$$
we should have equalities everywhere and so $a_0,a_2,\ldots,a_{r-1}$ are equal to certain $k$ and $a_1,a_3,\ldots,a_{r}$ are equal to $m-k$. Substituting these values to the cited formula we see that the corresponding term equals ${m\choose k}^{r+1}$.
$*)$ Well, the usual Taylor formula
$$\left[\prod x_i^{c_i}\right]F=\frac1{\prod c_i!}\left(\prod \left(\frac{\partial}{\partial x_i}\right)^{c_i}\right)F(0,\ldots,0)$$
may be considered as a partial case of CN formula for the multisets of the form $\{0,0,\ldots,0\}$. This formula of course also allows to get the necessary coefficient (if we differentiate $(y_i+y_{i+1})^m$ $k_i$ times w.r.t. $y_i$ and $m-k_i$ times w.r.t. $y_{i+1}$, we must have $k_i+(m-k_{i-1})=m$ for all $i$, thus $k_0=\ldots=k_r=:k$, and we get ${m\choose k}^{r+1}$.)
