Let $f : X to Y$ be a morphism of schemes and $x mapsto y$ be factors at which $f$ is easy of relative dimension $N$.
Then does the map of stalks induce $hat{mathscr{O}_{Y, y}} [[T_1, dots, T_N]] cong hat{mathscr{O}_{X, x}}$?

For $N = 0$, that is customary.
(e.g., see 4.3.26 of Liu’s “Algebraic geometry and…”.)
For basic $N$, since there exists an open neighbour $U$ of $x$ such that $f|_U$ is the composition of etale $U to mathbb{A}_Y^N$ and the projection $mathbb{A}_Y^N to Y$, we might assume that $X = mathbb{A}_Y^N$.
So it suffices to point out :

Let A be a neighborhood ring with the maximal superb $mathfrak{m}$, and $mathfrak{n} = mathfrak{m} + (T_1, dots, T_N).$
Then $hat{A}[[T_1, dots, T_N]] cong A[T_1, dots, T_N]^{hat{}}$
(The completion with respect to $mathfrak{n}$)

How can I present it?

And it appears that evidently the converse holds.
If that is true, please counsel me its references.