This is more or less the proof in Matsumura, Theorem 8.12. Let $\mathfrak{m} := (x_{1},\dotsc,x_{n}) \subset A[x_{1},\dotsc,x_{n}]$ be the ideal. Note that $$ \textstyle \hat{A} = \varprojlim_{\ell \in \mathbb{N}} A/I^{\ell} $$ and $$ \textstyle A[[x_{1},\dotsc,x_{n}]]/(x_{1}-a_{1},\dotsc,x_{n}-a_{n}) \simeq \varprojlim_{\ell \in \mathbb{N}} A[x_{1},\dotsc,x_{n}]/(\mathfrak{m}^{\ell},x_{1}-a_{1},\dotsc,x_{n}-a_{n}) $$ where in the above isomorphism we use that $A$ is Noetherian, show that the $A$-algebra map $$ A[x_{1},\dotsc,x_{n}]/(\mathfrak{m}^{\ell},x_{1}-a_{1},\dotsc,x_{n}-a_{n}) \to A/I^{\ell} $$ sending $x_{i} \mapsto a_{i}$ is an isomorphism, then take projective limit.
The map $\varphi : A[x_{1},\dotsc,x_{n}] \to A/I^{\ell}$ sending $x_{i} \mapsto a_{i}$ is surjective, and $(\mathfrak{m}^{\ell},x_{1}-a_{1},\dotsc,x_{n}-a_{n}) \subseteq \ker \varphi$. Suppose $f(x_{1},\dotsc,x_{n}) \in \ker \varphi$. Reducing modulo $x_{1}-a_{1},\dotsc,x_{n}-a_{n}$, we may assume that $f = a \in A$ is a constant; then $a \in \ker\varphi$ means $a \in I^{\ell}$, namely we can write $a = \sum_{\mathbf{e}} c_{\mathbf{e}}a_{1}^{e_{1}} \dotsb a_{n}^{e_{n}}$ where $\mathbf{e} = (e_{1},\dotsc,e_{n})$ ranges over elements of $(\mathbb{Z}_{\ge 0})^{\oplus n}$ such that $e_{1} + \dotsb + e_{n} = \ell$. Then $a$ is equivalent modulo $x_{1}-a_{1},\dotsc,x_{n}-a_{n}$ to $\sum_{\mathbf{e}} c_{\mathbf{e}}x_{1}^{e_{1}} \dotsb x_{n}^{e_{n}}$, which lies in $\mathfrak{m}^{\ell}$.