Assumption in Proof that Every $p$-Group is Solvable

Question

I found this proof here

My question is with fourth bullet point. It says,

Note that $G_1 = Z(G)$ is normal in $G$. . . .So the quotient group $G/G_1$ is defined and furthermore $|G/G_1| = |G|\; / \; |G_1| = \frac{p^k}{p}.$

I know that $G_1 = Z(G)$ is not trivial, but how do we know that $|Z(G)| = p$ and not, say, $p^2$?