I can't find a reference for this fact, so I fear it's false, but here's my proof of it.
Let $I = (f_1, ..., f_n)$ and $J = (f_1^{17}, ..., f_n^{17}).$ If $J$ is proper, then there's a maximal ideal $\mathfrak{m}$ containing $J$. Thus $f_i^{17} \in \mathfrak{m}$ for each $i,$ meaning $f_i\in \mathfrak{m}$ for each $i.$ Hence $\mathfrak{m}$ contains $1,$ meaning it is not actually maximal. Thus $J$ cannot be proper.
That's a correct proof.
You can also give more direct proofs, not using maximal ideals. Indeed, it suffices to prove the following, for $I$ an arbitrary ideal and $f\in R$ :
Suppose $1\in (f)+ I$. Then for any $n$, $1\in (f^n)+I$.
And a proof of this is rather easy: write $1 = fr + i, r\in R, i\in I$ and take the $n$th power : $1= 1^n = (fr+i)^n = f^nr^n +$ a bunch of terms, every one of which is a product of terms including at least one $i$, in particular, a bunch of terms in $I$.
So $1 = f^nr^n+ j, j\in I$ and so we are done.
