Why does this ideal contain a power of that ideal?

Question

I'm reading the proof of Prop. I-18 in Eisenbud-Harris The Geometry of Schemes and I'm unsure of a claim they make. Let me just give you some necessary info:

Let $R$ be a commutative unital ring and let $f_1, \dots, f_n$ be elements of $R$ which generate the whole ring i.e. $(f_1, \dots, f_n) = R$. Let $g\in R$ and suppose there exist nonnegative integers $k_i (1\leq i \leq n)$ with $f_i^{k_i} g = 0$. Let $N$ be the maximum of these integers, so that $f_i^N g = 0$ for all $i$. The authors seem to make the claim that the ideal $(f_1^N, \dots, f_n^N)$ contains some power of the ideal $(f_1, \dots, f_n)=R$, but it is not clear to me why it should. Wouldn't any power of $(f_1, \dots, f_n)$ contain certain mixed products of the $f_i$ not divisible by any $f_i^N$?

Here is the (near) original statement which I have just altered slightly to replace $g=h$ to $g=0$ (you can find the original on page 19 here):

This implies that $g$ is annihilated by a power of the ideal generated by all of the $f_a^N$ for some $N$. But this ideal contains a power of the ideal generated by all the $f_a$, which is the unit ideal. Hence $g=0$ in $R$.

Answer 1

The power $P=nN-n+1$ answers the question: indeed if all powers but one, say $k_i$, are $<N$, i.e. $k_j\le N-1$, then $$k_i=P-\!\!\!\sum_{\substack{1\le j\le n\\j\ne i}}k_j\ge nN-n+1-(n-1)(N-1)=N.$$