f_lm_l) = 0$. By the Leibniz rule, $$\partial^j(\sum_{i=1}^l f_im_i) = \sum_{i=1}^l (\partial^j(f_i)) m_i,$$ as all other terms have a factor of $\partial^{\alpha}(m_i)$ for some $\alpha > 0$ and some $i$ and hence vanish since $m_i \in M_*$. Multiplying by a harmless constant, we see that $\frac{1}{j!} \sum_{i=1}^l (\partial^j(f_i)) m_i = 0$. By our assumption, $\sum_{i=1}^l (\frac{1}{j!}\partial^j(f_i))_0 m_i = 0$, but this sum is nothing but $\sum_{i=1}^l f_{i,j}m_i$. We have thus reduced ourselves to the case $j=0$.

Let $m = \sum_{i=1}^l f_{i,0} m_i$, and let $E$ be the $R$-submodule of $M$ generated by $\{m_1, \ldots, m_l\}$. We claim that $m \in x_n^q E$ for all $q \geq 1$. As $E$ is a finitely generated $R$-module and $R$ is a Noetherian local domain (whose maximal ideal contains $x_n$), this will imply that $m = 0$ [@matsumura Thm. 8.10(ii)], as desired. We prove $m \in x_n^q E$ for all $q$ by induction on $q$. For the base case, $q=1$, consider the difference $\sum_{i=1}^l f_i m_i - \sum_{i=1}^l f_{i,0}m_i$. On the one hand, by definition, this difference belongs to $x_nE$, as we have removed all terms which *a priori* may lack an $x_n$ factor. On the other hand, the first term is $0$ by 