What I have done so far:
Let $(G,*)$ be a finite non-trivial group with $Z(G) = [G,G]$, that is, $G/Z(G)$ is Abelian.
Define $n := \exp(G/Z(G))$ and $s := n-1$ and $m := \exp(Z(G))$.
Since $\exp(G/Z(G)) = n$, it follows $\forall x \in G: x^n \in Z(G)$ and $\forall x \in G\ \exists z \in Z(G): x^n = z \wedge x = x^{-(n-1)}*z = x^{-s}*z$.
From $Z(G) = [G,G]$ it follows that every commutator of $G$ commutes with every other element in $G$.
Let $x,y \in G$.
$[x,y] = x*y*x^{-1}*y^{-1} = (x^{-s}*z)*y*(x^{-s}*z)^{-1}*y^{-1} = x^{-s}*z*y*z^{-1}*x^{s}*y^{-1} = x^{-s}*y*x^{s}*y^{-1} = [x^{-s},y] = [x,y]^{-s}$
So $[x,y]^n = [x,y]^{1+s} = 1$.
We know $\forall w \in G'\ \exists r \in \mathbb N_+\ \exists a_1,...,a_r,b_1,...,b_r \in G\ \exists \ell_1,...,\ell_r \in \mathbb Z: w = [a_1,b_1]^{\ell_1}*...*[a_r,b_r]^{\ell_r}$.
With $Z(G) = [G,G]$ we infer $w^n = ([a_1,b_1]^{\ell_1}*...*[a_r,b_r]^{\ell_r})^n = ([a_1,b_1]^{\ell_1})^n*...*([a_r,b_r]^{\ell_r})^n = ([a_1,b_1]^n)^{\ell_1}*...*([a_r,b_r]^n)^{\ell_r} = 1^{\ell_1}*...*1^{\ell_r} = 1$
So $\forall w \in G': ord(w) \leq n$ and $\exp(Z(G)) = m \leq n = \exp(G/Z(G))$.
We assume $n > 1$ and $G$ is a $p$-group, that is, $\exists n_0 \in \mathbb N_+: n = p^{n_0}$.
Let $x \in G\setminus Z(G)$ such that $ord(x Z(G)) = p^{n_0} = \exp(G/Z(G))$.
By choice of $x$ and definition of $n$ it follows $\forall \ell \in \mathbb N_+: (\ell < n_0) \supset (x^{p^\ell} \notin Z(G))$, that is, $\exists y \in G: [y,x^{p^{n_0-1}}] \not= 1$.
Note $\forall \ell \in \mathbb N_+: [y,x^{p^{\ell}}] = [y,x]^{p^{\ell}}$ because every commutator in $G$ commutes with every element of $G$.
Moreover, $[y,x]^{p^{n_0}} = [y,x^{p^{n_0}}] = 1$ because $x^{p^{n_0}} \in Z(G)$. Thus, $ord([y,x]) \leq p^{n_0}$.
If $ord([y,x]) = p^{n_0} = n$, then $n = ord([y,x]) \leq \exp(G') = \exp(Z(G)) = m \leq n$.
Question: Why is $ord([y,x]) < p^{n_0}$ impossible? This last step eludes me. Thank you for your thoughts!!
By assumption $[y,x]^{p^{n_0-1}} = [y,x^{p^{n_0-1}}] \ne 1$. Since $o([y,x])$ is a power of $p$. it must be greater than $p^{n_0-1}$, so it must be $p^{n_0}$.