2.52:

For any group G, the commutator subgroup [G,G] is equal to the set of all elements of G of the form a₁ ... a_n a₁^-1 ... a_n^-1, for n at least two.

Firstly, the identity element of G can be expressed in this form, by taking all a_i equal to 1. Secondly, the inverse of the commutator [x,y] is equal to the commutator [y,x], so that [G,G] consists of all finite products of commutators (i.e., there is no need to use inverses of commutators in such an expression).

We extrapolate from the hint to show that any finite product [a₁,b₁] ... [a_n,b_n] of commutators can be expressed in the desired form. Namely,

[a₁,b₁] ... [a_n,b_n] = a₁(b₁ a₁^-1)b₁^-1a₂(b₂ a₂^-1)b₂^-1 ... a_n(b_n a_n^-1)b_n^-1 a₁^-1(a₁ b₁^-1)b₁a₂^-1(a₂ b₂^-1)b₂ ... a_n^-1(a_n b_n^-1)b_n, which equals

a₁(b₁ a₁^-1)b₁^-1a₂(b₂ a₂^-1)b₂^-1 ... a_n(b_n a_n^-1)b_n^-1 a₁^-1(b₁ a₁^-1)^-1b₁a₂^-1(b₂ a₂^-1)^-1b₂ ... a_n^-1(b_n a_n^-1)^-1b_n,

and thus has the desired form.

It is also necessary to show that every element of the form a₁ ... a_n a₁^-1 ... a_n^-1 lies in the commutator subgroup [G,G]. A quick way to see this is to note that, in the quotient G/[G,G], the image of that element is equal to 1, since G/[G,G] is abelian. This implies the original element in G lies in the kernel of the projection G ---> G/[G,G], which is [G,G].