Here $p$ is a prime number and $n>1$.
I know the following facts:
$p\mathbb{Z}/p^n\mathbb{Z}$ is a prime (hence maximal) ideal of $\mathbb{Z}/p^n\mathbb{Z}$.
$\mathbb{Z}/p\mathbb{Z}$ is a field.
So my conjecture is that $(\mathbb{Z}/p^n\mathbb{Z})/(p\mathbb{Z}/p^n\mathbb{Z})\cong\mathbb{Z}/p\mathbb{Z}$. However, is seems strange to me that it would then not depend on $n$, which makes me suspect this is not true. I would have expected this quotient to be isomorphic to $\mathbb{Z}/p^{n-1}\mathbb{Z}$, but this is not a field in general. My only other guess (and this one is a shot in the dark) would be the unique finite field of order $p^{n-1}$.
Expanding on carmichael561's comment: explicitly, consider the projection
$$ \pi : m \in \mathbb{Z} \mapsto [m] \in \mathbb{Z}/p\mathbb{Z}\newcommand{\Z}{\mathbb{Z}}. $$
Since $p^n\Z \subset p\Z$, this map factorizes through $\Z/p^n\Z$ yielding a morphism
$$ \widehat{\pi} : [m] \in \Z/p^n\Z \mapsto [m] \in \Z/p\Z. $$
which is an epi (because $\pi$ is). Now, by the first isomorphism theorem we obtain
$$ \Z/p\Z \simeq (\Z/p^n\Z) / \ker \widehat{\pi}, $$
so it would suffice to show that $\ker \widehat{\pi} = p\Z/p^n\Z$. In effect, an element $[m] \in \Z/p^n\Z$ is in the kernel of $\widehat{\pi}$ if the class of $m$ in $\Z/p\Z$ is zero, and this occurs precisely when $m = pk$ for some integer $k \in \Z$. Since $[pk] = p[k]$, this is to say $[m] \in p\Z/p^n\Z$.
Finally, an unimportant "intuitive" remark: the ideal $p\Z$ is "large", in the sense that $p^k\Z \subset p^l\Z$ when $k> l$. The same holds for $p^k\Z/p^n\Z$ and $p^l\Z/p^n\Z$. Hence dividing by $p\Z/p^n\Z$ should produce a "small" ring (in this case, a field).