**Proof plan.** Let $G$ act on itself by conjugation and apply the class equation; the $p$-divisibility of all non-singleton orbit sizes forces the center to be non-trivial. **Step 1: The class equation.** The conjugacy classes partition $G$. An element $g$ has conjugacy class $\{e\}$ (size $1$) if and only if $g$ commutes with all of $G$, i.e. $g \in Z(G)$. All other conjugacy classes have size $> 1$. By the [Orbit-Stabilizer Theorem](/theorems/845), each conjugacy class size divides $|G| = p^n$, so every size is a power of $p$. Non-singleton classes thus have size divisible by $p$. The class equation gives \begin{align*} |G| = |Z(G)| + \sum_{\substack{g \notin Z(G) \\ \text{one per class}}} |G : C_G(g)|. \end{align*} **Step 2: $p$ divides $|Z(G)|$.** Every term $|G : C_G(g)|$ in the sum (for $g \notin Z(G)$) is divisible by $p$. Since $|G| = p^n$ is also divisible by $p$, we conclude $p \mid |Z(G)|$. **Step 3: $Z(G)$ is non-trivial.** Since $p \mid |Z(G)|$ and $p \geq 2$, we have $|Z(G)| \geq 2$, so there exists $x \neq e$ with $x \in Z(G)$. $\square$