Finite number of elements generating the unit ideal of a commutative ring
Solution 1:
Since $(f_1,\ldots, f_r) = A$, there exist elements $a_i\in A$ such that $a_1f_1 + \cdots + a_rf_r = 1$. We then get that $$1 = (a_1f_1 + \cdots + a_rf_r)^{rn}.$$ Expanding the right hand side out explicitly shows that it is in the ideal $(f_1^n,\ldots, f_r^n)$. Therefore $(f_1^n,\ldots, f_r^n) = (1) = A$.
Solution 2:
Alternatively, clearly $(f_1,\dots,f_r) \subset \sqrt{(f^{n_1}_1,\dots, f^{n_r}_r)}$. Thus, $\sqrt{(f^{n_1}_1,\dots, f^{n_r}_r)} = A$, and so $1 \in (f^{n_1}_1,\dots, f^{n_r}_r)$.