Let $A$ be a Noetherian ring and $ q \in\mathrm{Spec} (A)$. Then $q^{(n)} A_q=q^{n}A_q$, where $q^{(n)}= \lbrace a \in A \mid \exists d \in A \setminus q\text{ such that }da \in q^n \rbrace$ and $A_q= S^{-1}A$, $S=A \setminus q$.
My proof is this:
The inclusion $\supseteq$ is obvious.
For inclusion $\subseteq$ : I considered $\sum_{i=1}^{p} x_i \frac{a_i}{b_i} \in q^{(n)}A_q $.
For each $i$ there exists $d_i \in A \setminus q$ such that $d_ix_i \in q^n$. We have that $d=d_1...d_p \notin q$ and $$(\sum_{i=1}^{p} x_i \frac{a_i}{b_i}) d_1...d_p \in q^{n}A_q \Rightarrow (\sum_{i=1}^{p} x_i \frac{a_i}{b_i}) d_1...d_p =\sum_{j=1}^{r} y_j \frac{t_j}{s_j} \Rightarrow \sum_{i=1}^{p} x_i \frac{a_i}{b_i} =\sum_{j=1}^{r} y_j \frac{t_j}{ds_j} \in q^n A_q$$
It is right ?
We have $q^{(n)}=(q^n)^{ec}$, so $q^{(n)}A_q=(q^{(n)})^e=(q^n)^{ece}=(q^n)^e=q^nA_q$. (See Atiyah and Macdonald, Introduction to Commutative Algebra, Proposition 1.17 and the section "Extension and Contraction".)
Regarding to your proof: I'd start with $a\in q^{(n)}$, and $s\in A\setminus q$ (this is actually not necessary), and try to show that $\frac as\in q^nA_q$. Since there is $d\in A\setminus q$ such that $da\in q^n$ we are almost done: $\frac as=\frac{da}{ds}\in q^nA_q$. (In particular, this shows that your proof is longer than enough, but correct.)