Define $G$ to act on $H$ by conjugation; that is, for $g\in G$ $h\in H$ define $$g \cdot h = ghg^{-1}$$ Note that $g\cdot h\in H$ since $H\triangleleft G$ This is easily seen to be a well-defined group action.

Now, the set of invariants of $H$ under this action are $$G_H=\{h \in H \ \lvert \ g \cdot h = h \ \forall g\in G\} = \{h \in H \ \lvert \ ghg^{-1} = h \ \forall g\in G\}=H \cap Z(G)$$

The class equation theorem states that $$\lvert H\rvert = \lvert G_H\rvert + \sum_{i=1}^{r}[G:G_{x_{i}}]$$ where the $G_{x_{i}}$ are proper subgroups of $G$ and thus that $$\lvert G_H\rvert = \lvert H\rvert - \sum_{i=1}^{r}[G:G_{x_{i}}]$$

We now use elementary group theory to show that $p$ divides each term on the right, and conclude as a result that $p$ divides $\lvert G_H\rvert$ so that $G_H=H\cap Z(G)$ cannot be trivial.

As $G$ is a nontrivial finite $p$ group, it is obvious from Cauchy's theorem that $|G|=p^{n}$ for $n>0$ Since $H$ and the $G_{x_i}$ are subgroups of $G$ each either is trivial or has order a power of $p$ by Lagrange's theorem. Since $H$ is nontrivial, its order is a nonzero power of $p$ Since each $G_{x_i}$ is a proper subgroup of $G$ and has order a power of $p$ it follows that $[G:G_{x_i}]$ also has order a nonzero power of $p$