Isomorphism of quotients of powers of maximal ideals Let $R$ be an integral domain, and $\mathfrak{m}$ a maximal ideal of $R$.  Let $R_\mathfrak{m}$ denote the ring localized at $\mathfrak{m}$, and let $\mathfrak{m}_\mathfrak{m} = \mathfrak{m}R_\mathfrak{m}$ denote the maximal ideal of $R_\mathfrak{m}$.
Then we have an isomorphism $$R/\mathfrak{m}^n \cong R_\mathfrak{m}/\mathfrak{m}_\mathfrak{m}^n,$$ for $n \geq 1$.  (See for example Neukirch, Algebraic Number Theory, pg. 66).
My question:
Can this result extend to an isomorphism
$$\mathfrak{m}^k/\mathfrak{m}^n \cong \mathfrak{m}_\mathfrak{m}^k/\mathfrak{m}_\mathfrak{m}^n,$$
for $1 \leq k \leq n$?  It would seem the proof goes through, but I wanted to check with the experts, since I hadn't seen this result explicitly written down before.
2nd question (added later):
Is the "integral domain" condition really necessary to either Neukirch's result or the other one I suggested?
 A: I suggest a different approach in order to prove that $\mathfrak{m}^k/\mathfrak{m}^n \simeq \mathfrak{m}_\mathfrak{m}^k/\mathfrak{m}_\mathfrak{m}^n$ which also contains the case $k=0$. 
Obviously $\mathfrak{m}_\mathfrak{m}^k/\mathfrak{m}_\mathfrak{m}^n$ is isomorphic to $S^{-1}(\mathfrak{m}^k/\mathfrak{m}^n)$, where $S=R-\mathfrak m$. If we set $M=\mathfrak{m}^k/\mathfrak{m}^n$ we have to prove that $M\simeq S^{-1}M$. But $\mathfrak m^tM=0$ for $t=n-k\ge 1$, and thus $M$ is in fact an $\overline R=R/\mathfrak m^t$-module. Now we can replace $S$ by $\overline S=\overline R-\overline{\mathfrak m}$ and let us see what we have now: 

$\overline R$ is a local ring with maximal ideal $\overline{\mathfrak m}$, $\overline S=\overline R-\overline{\mathfrak m}$, and $M$ an $\overline R$-module. We want to prove that $M\simeq\overline{S}^{-1}M$. 

But there is nothing to prove once we note that $\overline S=U(\overline R)$. 
A: Well I found this result in Milne's Commutative Algebra notes, Prop 5.8, and the ring need not be either Noetherian or an integral domain.
A: Yes, $\mathfrak{m}^k/\mathfrak{m}^n \cong \mathfrak{m}_\mathfrak{m}^k/\mathfrak{m}_\mathfrak{m}^n$.
This immediately results from the commutative diagram 
$$\begin{array}
00 & \longrightarrow \mathfrak{m}^k/\mathfrak{m}^n\longrightarrow & R/\mathfrak{m}^n & \longrightarrow & R/\mathfrak{m}^k& \longrightarrow 0\\
& \quad \quad \downarrow & \downarrow \cong  & & \downarrow \cong\\
0 &\longrightarrow  \mathfrak{m}_\mathfrak{m}^k/\mathfrak{m}_\mathfrak{m}^n\longrightarrow   & R_\mathfrak{m}/\mathfrak{m}_\mathfrak{m}^n & \longrightarrow & R_\mathfrak{m}/\mathfrak{m}_\mathfrak{m}^k& \longrightarrow 0  
\end{array}
$$  Since you know that the the two rightmost vertical maps are isomorphisms, the leftmost vertical map is also an isomorphism, by the snake lemma (say).
