Finite number of elements generating the unit ideal of a commutative ring

Let $A$ be a commutative ring with $1$.
Let $f_1,\dots,f_r$ be elements of $A$.
Suppose $A = (f_1,\dots,f_r)$.
Let $n > 1$ be an integer.
Can we prove that $A = (f_1^n,\dots,f_r^n)$ without using axiom of choice?

EDIT
It is easy to prove $A = (f_1^n,\dots,f_r^n)$ if we use axiom of choice.
Suppose $A≠(f_1^n,…,f_r^n)$. By Zorn’s lemma(which is equivalent to axiom of choice), there exists a prime ideal $P$ such that $(f_1^n,…,f_r^n) \subset P$. Since $f_i^n \in P$, $f_i \in P$ for all $i$.
Hence $A = (f_1, \dots,f_n) \subset P$.
This is a contradiction.

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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 $1 \in (f^{n_1}_1,\dots, f^{n_r}_r)$.

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 $(f_1^n,\ldots, f_r^n) = (1) = A$.