Trouble understanding Eisenbud Exercise 2.19a

For any multiplicatively closed subset $S$ of $R$, there is a canonical mapping $\,\varphi\colon M\rightarrow S^{-1}M$ defined by $\,\varphi(m)=\dfrac m1$. By definition of a module of fractions, $\varphi(m)=0\iff \exists s\in S,\ sm=0$.

Here we have multiplicatively closed subsets $S_i=\bigl\{1, f_i, f_i^2,\dots, f_i^k,\dots\bigr\}$, and $m$ maps to $0$ in $M[f_i^{-1}]$ if and only if there exists a power $k_i$ of $f_i$ such that $f_i^{k_i}m=0$.

Let $K=k_1+\dots+k_n$; write $1=\lambda_1 f_1+\dots+\lambda_nf_n$. Then by the multinomial formula, $$1=(\lambda_1 f_1+\dots+\lambda_nf_n)^K=\sum_{r_1+\dots+r_n=K} \frac{N!}{r_1!\dotsm r_n!}(\lambda_1 f_1)^{r_1}\dots(\lambda_n f_n)^{r_n}. $$ Each term in this sum kills $m$. Indeed since $r_1+\dots+r_n=k_1+\dots+k_n$, at least $1$ $r_i$ is $\geq k_i$. Thus $1\cdot m = m= 0$.

If $\psi_i(m) = 0$, then you know that $f_i^{n_i} m = 0$ for some $n_i \geq 0$. If you could express $1 = \sum_{i=1}^n a_i f_i^{n_i}$ with $a_i \in R$, then you'd be done, for you would know that $m = \sum_{i=1}^n a_i f_i^{n_i} m = 0$. But you know that $\langle f_1, \dots, f_n \rangle = R$, so $1 = c_1 f_1 + \dots + c_n f_n$. Raise each side of this equation to the $\sum_{i=1}^n n_i$ power and notice that when you expand by the binomial theorem, each term contains a factor of the form $f_i^{n_i}$ for at least one $i$.