This is Exercise 3.36 of Roman's "Fundamentals of Group Theory: an Advanced Approach". According to Approach0, it is new to MSE.
The Details:
Neither semidirect products nor presentations are covered in the book so far.
Definition: The commutator subgroup or derived subgroup $G'$ of a group $G$ is the subgroup of $G$ generated by all commutators $$[a,b]:=aba^{-1}b^{-1}$$ (for $a,b\in G$). A group $G$ is perfect if $G=G'$.
The Question:
If $G$ is a finite group of odd order, then the product of all its elements, in any order, is an element of the commutator subgroup $G'$.
Thoughts:
If $G$ is abelian, then $$\prod_{g\in G}g=e\in G'$$ trivially.
If $G$ is perfect, there is nothing to prove.
By Lagrange's Theorem, $G$ has no involutions, so there are no elements $x\in G$ such that $x=x^{-1}$.
Putting the identity to one side, then, elements come in pairs: $y$ and $y^{-1}$. Let's separate $G$ into any $\frac{n-1}{2}$ elements $g$ in a set $P$; and their inverses $g^{-1}$, a set $N$. Note that $P\cap N=\varnothing$. Now one product of all elements of $G$ is of the form
$$e\left(\prod_{g\in P}gg^{-1}\right)=e\in G',$$
for any order of the elements of $P$.
So much for the theoretical approach . . .
Consider when $G=\Bbb Z_7\rtimes \Bbb Z_3$, given by the presentation
$$\langle a,b\mid a^7, b^3, bab^{-1}=a^4\rangle,$$
which I believe to be the smallest, nonabelian group of odd order, according to group names.
Here $G'\cong \Bbb Z_7$, again, according to group names.
The elements of $G$ are of the form $a^ib^j$ for $i\in \{0,\dots, 6\}$ and $j\in\{0,1,2\}$. Again, taking the identity aside, there is, essentially, up to $20!$ different products of all elements of $G$; there might be subtleties that lower that upper bound, but still: it's a lot: that's what I'm getting at. Consider $[a^{\alpha_1}b^{\beta_1}, a^{\alpha_2}b^{\beta_2}].$ This is looks like it would be messy to compute in abstraction. There ought to be a slick way of doing this.
Please help :)
Hint: $G/G'$ is abelian, the image of the product of elements in any order by the quotient map $G\rightarrow G/G'$ is the neutral element of $G/G'$, a fact that you have noticed