Every $p$-group is nilpotent

Let \(G\) be a finite \(p\)-group. Every nontrivial finite \(p\)-group has nontrivial center: by the class equation, \[ |G|=|Z(G)|+\sum_i [G:C_G(x_i)], \] and every summand in the sum is divisible by \(p\), hence \(p\mid |Z(G)|\).

Now form the upper central series \[ 1=Z_0(G)\le Z_1(G)\le Z_2(G)\le\cdots, \qquad Z_{i+1}(G)/Z_i(G)=Z(G/Z_i(G)). \] If \(Z_i(G)\ne G\), then \(G/Z_i(G)\) is a nontrivial finite \(p\)-group, so its center is nontrivial. Hence \[ Z_i(G)<Z_{i+1}(G). \] Each strict inclusion multiplies the order by at least \(p\), so after finitely many steps the series reaches \(G\). Therefore \(G\) is nilpotent.