 Solution

Use induction on the order of the finite p-group. The trivial group is nilpotent. For $G\ne 1$, the permitted center theorem gives $Z=Z(G)\ne 1$, so the quotient group $G/Z$ is a smaller p-group. By induction it has a central seriesLet $H_i$ be the inverse image of ${\overline{H}}_i$ under the quotient map, so $H_0=Z$ and $H_s=G$. Centrality in the quotient gives $[H_i,G]\le H_{i-1}$ for $i\ge 1$: the image of every such group commutator lies in ${\overline{H}}_{i-1}$. Also $[Z,G]=1$. Thereforeis a central series of $G$. This proves every finite p-group is nilpotent, including the case $Z=G$, when the quotient is trivial and the series is simply $1\le G$.