Let $G$ be a group, $G^{(n)}$ be its $n$-derived group (i.e $G^{(n+1)}=[G^{(n)},G^{(n)}]$ and $G^{(0)}=G$) and $G^n$ be its $n$-term in the lower central series (i.e $G^{n+1}=[G^n,G]$ and $G^0=G$).
Are the following identities true? $$G^{(2)}=<[[x_1,x_2],[x_3,x_4]]\mid x_1,x_2,x_3,x_4\in G>$$ $$G^{2}=<[[x_1,x_2],x_3]\mid x_1,x_2,x_3\in G>$$
I am also interested in the analogous statements for $n$ bigger than 2.
My motivation for this is that I want to write the statements $G^n=\{e\}$ and $G^{(n)}=\{e\}$ as first order formulas in order to use them in a problem from Model theory, and if I have that explicit set of generators for them then is easy to do it.
I think the answer is yes. An element of $G^{(2)}$ is a product of commutators of elements of $G^{(1)}$, and these elements of $G^{(1)}$ are themselves products of commutators of elements of $G$.
So an element of $G^{(2)}$ is a product of elements of the form $[u_1\cdots u_m,v_1 \cdots v_n]$, where the $u_i$ and $v_j$ are commutators. But by the commutator formulae, this is a product of elements of form $[u_i,v_j]^g$ for $g \in G$, which is $[[a,b],[c,d]]^g = [[a^g,b^g],[c^g,d^g]]$ for some $a,b,c,d, \in G$, and that lies in your putative generating set.
A similar argument works for $G^2$. I have not thought about temrs with $n>2$ and I will leave that to you.