Isomorphism of quotients of powers of maximal ideals

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).

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 $\overline S=U(\overline R)$.

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.