Express the identity element as a linear combination of $N^{th}$ powers of generators in a ring 
Let $R$ be a commutative ring and suppose $f_1,\ldots,f_k \in R$. If $(f_1,\ldots,f_k)=R$, show that there exist $r_i\in R$ such that $$\sum_i r_i{f_i}^N=1$$ for every positive integer $N$.

I tried using induction to prove this claim. For $N=1$, the statement is obviously true since $f_i$ are the generators of $R$. I know that if $f_i$ generate the whole ring then every element in $R$ can be expressed as $\pm \prod{f_i}^{r_{1i}} \pm \prod{f_i}^{r_{2i}} \ldots \pm \prod{f_i}^{r_{ni}}$ but I don't know how to use this to complete the induction. Any thoughts?

Since we know that $(f_1,\ldots,f_k)=R$, we know that $1\in (f_1,\ldots,f_k)$, so there exist polynomials $g_i\in R$ such that $\sum{f_ig_i}=1$.
Now raise this equation to the $Nk$th power to get $(\sum{f_ig_i})^{Nk}=1$. If you expand this out using the multinomial theorem, we see that every term contains at least one $f_i$ with a power of $N$ or greater. If this were not the case, such a term would be (without the coefficient) $\prod_{i=1}^k f_i^{a_i} $ so that the number of $f_i$ showing up in this term are strictly less than $ Nk$, a contradiction. This proves your claim.
