$p^n$ power torsion and $p^{n+1}$ power torsion defined over the same field Let $E$ be an elliptic curve over $\mathbb{Q}$ with no rational $p$-torsion, and with good reduction over $\mathbb{Q}$. (I'm not sure if the good reduction is really necessary here.) For each $n$, we can form the field generated by the coordinates of the $p^n$ torsion points of $E$, $\mathbb{Q}(E[p^n])$. My question is: is this field ever the same for $n$ as for $n+1$?
In the language of Galois representations, the $p$-adic representation on the Tate module $T_p(E)$ has reductions $\rho_n: G_\mathbb{Q} \to GL_2(\mathbb{Z}_p) \to GL_2(\mathbb{Z}/p^n\mathbb{Z})$ for each $n$ (fixing a basis of $T_p(E)$ for simplicity). Furthermore, $\rho_n = \pi_n \circ \rho_{n+1}$ where $\pi_n$ is the projection map $GL_2(\mathbb{Z}/p^{n+1} \mathbb{Z}) \to GL_2(\mathbb{Z}/p^n\mathbb{Z})$. My question could also be phrased as: is it possible that $\ker \pi_n \cap \textrm{im } \rho_{n+1}$ is trivial?
 A: tl;dr: yes for $n=1$ and $p=2$ otherwise no.
Let $L_n:=\mathbf{Q}(E[p^n])$ and $K_n:=\mathbf{Q}(\zeta_{p^n})$. We have maps:
$$\mathrm{Gal}(L_n/\mathbf{Q}) \subseteq \mathrm{GL}_2(\mathbf{Z}/p^n \mathbf{Z}) \stackrel{\mathrm{det}}{\rightarrow}
(\mathbf{Z}/p^n \mathbf{Z})^{\times} = \mathrm{Gal}(K_n/\mathbf{Q}),$$
and a corresponding inclusion $K_n \subset L_n$. If $L_n = L_{n+1}$, then there is a surjection:
$$\mathrm{Gal}(L_n/\mathbf{Q}) \rightarrow \mathrm{Gal}(K_{n+1}/\mathbf{Q})
= (\mathbf{Z}/p^{n+1} \mathbf{Z})^{\times}.$$
For $m \ge 1$, the group $\mathrm{Gal}(L_{m+1}/L_m)$ is a (possibly trivial) elementary group of exponent $p$. Hence it follows from the existence of the surjection above that there are also exist surjections
$$\mathrm{Gal}(L_m/\mathbf{Q})  \rightarrow (\mathbf{Z}/p^{m+1} \mathbf{Z})^{\times}$$
for every $m \le n$, and in particular for $m = 1$. We show that  for $p > 2$ there are no groups  $G \subset \mathrm{GL}_2(\mathbf{F}_p)$  with this property and such that the determinant map is surjective.
Since $G$ has order dividing $p$, it either contains $\mathrm{SL}_2(\mathbf{F}_p)$ (and thus is $\mathrm{GL}_2(\mathbf{F}_p)$ by the determinant assumption) or is contained in a Borel $B$.
In the first case, there are no surjective maps to a cyclic group of order $p$ when $p >2$, because $\mathrm{SL}_2(\mathbf{F}_p)$ is simple up to the center of order $2$ when $p \ge 5$, and $G = \widetilde{S_4}$ when $p = 3$ and the abelianization has order $2$.
If $G \subset B$, we may assume after conjugation that $B$ consists of upper triangular matrices and $G$ contains the element
$$\gamma = \left( \begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix} \right).$$
If $\beta \in B$, then the commutator $[\beta,\gamma]$ is a non-trivial power of $\gamma$ unless $\beta$ lies in the group $H$ generated by $\gamma$ and by the diagonal matrices. In particular, the only subgroups of the Borel whose commutator subgroup does not contain an element of order $p$ (which would prevent the existence of a quotient of order $p$) lie inside a group  $H$ all of whose elements have determinant in $\mathbf{F}^{\times 2}_p$ which prevents the determinant map from being surjective.
This leaves the case $p = 2$. In this case, it can indeed happen that $\mathbf{Q}(E[2]) = \mathbf{Q}(E[4])$, for example when:
$$E:y^2 + x y + y = x^3 - x^2 + 4 x - 1.$$
It turns out that it is impossible that  $\mathbf{Q}(E[2^n]) = \mathbf{Q}(E[2^{n+1}])$ for larger $n$, but this is a theorem of Jeremy Rouse and David Zureick-Brown that depends on showing that certain modular curves have no rational points. See Remark 1.5 of https://arxiv.org/pdf/1402.5997.pdf
The example $E$ above has a surjective $2$-torsion representation. Finally, you have the assumption "with good reduction over $\mathbf{Q}$" but this doesn't make much sense because elliptic curves by definition are smooth. If you mean "good reduction over $\mathbf{Z}$" then there are no elliptic curves at all. If you mean "good reduction at $p$" then there are also no curves; the argument above shows that $\mathbf{Q}(\zeta_{p^2}) \subset \mathbf{Q}(E[p])$, and this prevents $E[p]/\mathbf{Z}_p$ arising from  a finite flat group scheme by Fontaine's bounds. 
