Let $k,m,n \in \mathbb{N}$ such that $2 \leq k+1 < n < m$ and $a_0, a_1, a_2, \dots, a_k \in \mathbb{C}$.
Let $g = a_0 + a_1X + \dots + a_kX^k + X^m$ and $f = a_0 + a_1X + \dots + a_kX^k + X^n$ and suppose $f$ divides $g$.
Show that $a_0 a_1 a_2 \cdots a_k = 0$.
This is what I tried:
If $f \mid g$ then $f \mid g - f = X^m - X^n = X^n(X^{m-n} - 1)$ So if $f$ has $0$ as a root then $a_0$ will be $0$ and hence the result. So if we assume $f$ doesn't have a $0$ root then $f \mid X^{m-n} - 1$, so $f$ contains only roots of unity.
Now I'm stuck.
As you have observed, $f|g\Leftrightarrow f|(g-f)=X^n(X^{m-n} - 1)\Leftrightarrow$ $\{\text{roots of $f$}\}\subseteq \{\text{roots of $X^n(X^{m-n} - 1)$}\}=\{0,\omega,\omega^2,\cdots,\omega^{m-n}=1\}$, where $\omega=e^{\frac{2\pi i}{m-n}}$. If $0$ is a root of $f$, then $a_0=0$. We are going to prove that if $0$ is not admitted as a root of $f$, then $a_1=0$.
In fact, if $f(0)\ne 0$, then $f=(X-x_1)\cdots(X-x_n)$, where $x_i=\omega^{b_i}$ for some $1\le b_i\le m-n$ for each $1\le i\le n$. The hypothesis that $a_{n-1}=0$ implies $x_1+\cdots+x_n=0$ \begin{equation*} \Rightarrow \overline{x_1+\cdots+x_n}=0 \end{equation*} \begin{equation*} \Rightarrow \frac{1}{x_1}+\cdots+\frac{1}{x_n}=0 \end{equation*} Writing the last equation as $\frac{{\rm numerator}}{x_1\cdots x_n}=0$, you will find the numerator is $a_1$ up to a $\pm$ sign, as desired.