 Solution

By Lagrange's theorem, a stabilizer subgroup of $G$ has order $p^j$ for some $0\le j\le n$. The orbit-stabilizer theorem therefore givesThus an orbit of a group action either has one element or has size divisible by $p$, even if the acted-on set is infinite.

Apply this to the conjugation action. Its singleton orbits of a group action are the elements of the centre of a group, while every other orbit has size divisible by $p$. Partitioning the finite group into these orbits gives the class equation, and henceSince the identity belongs to $Z(G)$, its order is positive and divisible by $p$. ConsequentlyNow suppose that $|Z(G)|=p^{n-1}$. It is then proper, so choose $x\notin Z(G)$. The centralizeris a subgroup: commuting with $x$ is preserved by multiplication and inverses. It contains $Z(G)$ and also $x$, so $|C_G(x)|>|Z(G)|=p^{n-1}$. By Lagrange's theorem, its order is a power of $p$ dividing $p^n$, forcing $C_G(x)=G$. But that would make $x$ central, a contradiction. Therefore