Coefficients in products and powers of large polynomials Let $f\in \mathbb{Z}[x_1,\dots,x_n]$ be a polynomial. I want to show that a certain monomial $m$ shows up with non-zero coefficient in the $r^{th}$ power of $f$. 
If you're lucky, you can do this as follows:
Regrade $\mathbb{Z}[x_1,\dots,x_n]$ such that $m$ appears with non-zero coefficient in the initial form $(\textrm{init}f)^r$. Then because $m$ has highest weight, no trailing terms of $f^{r}$ can cancel $m$. This means you can take a hard calculation with a long, complicated polynomial and turn it into an easy calculation with a short polynomial.
Sadly, $m$ may be interior in the Newton polytope of $f^r$, in which case the above trick has no hope of working.
I would be grateful for other techniques people use for showing that monomials show up with non-zero coefficent in powers (and products) of polynomials. 
Thanks!
(In fact, I really care about the situation $f\in \mathbb{F}_p[x_1,\dots,x_n]$ and $r = p-1$ for some prime $p$.)
 A: The coefficient of $\prod_{i=1}^nx^{k_i}$ in $f$ is the same thing as $(\partial_{(\bar{k})} f)|_{\bar{x}=\bar{0}}$, where $\partial_{(\bar{k})}=\prod_{i=1}^n\frac{\partial_{x_i}^{k_i}}{k_i!}$. You can take advantage of this to simplify calculations. As a small example, consider $$\begin{align}f(x,y,z) &= 1+x +y -2xy + 3xz^2 & m&=x^2y^2z^2\end{align}$$ with $r = 5$.
$$\begin{align}
(\partial_{(2,2,2)} f^5)|_{(x,y,z)=\bar{0}} & =\frac{1}{8}(\partial_{x}^2\partial_{y}^2\partial_{z}^2 f^5)|_{(x,y,z)=\bar{0}}\\
& =\frac{1}{8}(\partial_{x}^2\partial_{y}^2\partial_{z} 5f^4f_z)|_{(x,y,z)=\bar{0}}\\
& =\frac{1}{8}\partial_{x}^2\partial_{y}^2 (20f^3f_z^2+5f^4f_{zz})|_{(x,y,z)=\bar{0}}\\
\end{align}
$$
Now that we will never again differentiate by $z$, it is acceptable to evaluate at $z=0$. Let $g(x,y)=f(x,y,0)=1+x+y-2xy$. Note that $f_z(x,y,0)=0$ and $f_{zz}(x,y,0)=6x$, simplifying what we have so far. We are left with:
$$\begin{align}
(\partial_{(2,2,2)} f^5)|_{(x,y,z)=\bar{0}} & =\frac{1}{8}\partial_{x}^2\partial_{y}^2 (30xg^4)|_{(x,y)=\bar{0}}\\
& =\frac{1}{8}\partial_{x}^2\partial_{y} (120xg^3g_y)|_{(x,y)=\bar{0}}\\
& =\frac{1}{8}\partial_{x}^2(360xg^2g_y^2+120xg^3g_{yy})|_{(x,y)=\bar{0}}\\
& =\frac{1}{8}\partial_{x}^2(360xg^2g_y^2)|_{(x,y)=\bar{0}}\\
\end{align}
$$
where we used the fact that $g_{yy}=0$. As we will never again differentiate with respect to $y$... let $h(x)=g(x,0)=1+x$.
$$\begin{align}
(\partial_{(2,2,2)} f^5)|_{(x,y,z)=\bar{0}} & =\frac{1}{8}\partial_{x}^2(360xh^2(1-2x)^2)|_{(x)=\bar{0}}\\
& =\frac{1}{8}\partial_{x}^2(360x(1+x)^2(1-2x)^2)|_{(x)=\bar{0}}\\
& =\frac{1}{8}\partial_{x}^2(360x(1+2x+x^2)(1-4x+4x^2))|_{(x)=\bar{0}}\\
& =\frac{1}{8}\partial_{x}^2(360x(1+2x)(1-4x))|_{(x)=\bar{0}}\\
& =\frac{1}{8}\partial_{x}^2(360x(1-2x-8x^2))|_{(x)=\bar{0}}\\
& =\frac{1}{8}\partial_{x}^2(-720x^2)|_{(x)=\bar{0}}\\
& = \frac{1}{8}(-720)(2)\\
& = -180
\end{align}
$$
where in this final block we have stayed conscious of the degree to which we are differentiating and that we will evaluate at $0$ . 
I think this strategy (using derivatives and evaluating at $0$) will generally be more efficient than directly computing $f^r$. This should also be applicable to products of polynomials.
