The commutator subgroup $G'$ of $G$, is the set of all "long commutators" By definition, the commutator is the subgroup generated by all commutators, that is
$$G' = \langle\{aba^{-1}b^{-1}\mid a,b \in G\}\rangle$$
I'd like to prove that also
$$X = \{a_1a_2\cdots a_na_1^{-1}a_2^{-1}\cdots a_n^{-1} | a_i\in G, n\ge2\}$$
is the commutator subgroup. That is $X = G'$.
I already got one contention, as for $x \in G'$ 
$$ x = (a_1a_2a_1^{-1}a_2^{-1})\cdots(a_{2n-1}a_{2n}a_{2n-1}^{-1}a_{2n}^{-1})$$
as it is a finite product of commutators. Clearly, that's the same as
$$ x = (a_1a_2a_1^{-1}a_2^{-1})\cdots(a_{2n-1}a_{2n}a_{2n-1}^{-1}a_{2n}^{-1})(a_1^{-1}a_1\cdots a_{2n}^{-1}a_{2n})$$
By associativity, 
$$x = (a_1(a_2a_1^{-1})a_2^{-1} \cdots a_{2n-1}(a_{2n}a_{2n-1}^{-1})a_{2n}^{-1})(a_1^{-1}(a_1a_2^{-1})a_2\cdots a_{2n-1}^{-1}(a_{2n-1}a_{2n}^{-1})a_{2n})$$
Then $x \in X$ and $G' \subseteq X$.
Now I'm stuck with the other contention. It seemed easier at first, but now I'm not that sure. 
 A: This is a question in Rotman's Group Theory:

(P.Yff).
  For any group $G$, show that $G'$ is the subset of all “long commutators”:
  $$
    G'
  = \{
       a_1 a_2 \dotsm a_n a_1^{-1} a_2^{-1} \dotsm a_n^{-1}
    :
       \text{$a_i \in G$ and $n \geq 2$}
    \}.
$$
  (Hint (P.M. Weichsel).
  $$
    (a b a^{-1} b^{-1}) (c d c^{-1} d^{-1})
  =        a      (b a^{-1})
    b^{-1} c      (d c^{-1})
    d^{-1} a^{-1} (a b^{-1})
    b      c^{-1} (c d^{-1})
    d. )
$$
(Original image here.)


Here is an elementary proof by induction. 
$$a_1\cdots a_{n-1} a_n a_1^{-1}\cdots a_{n-1}^{-1}a_n^{-1}=(a_1\cdots a_{n-1})a_n (a_{n-1}\cdots a_1)^{-1} a_n^{-1}.$$
We insert the term $(a_{n-1}\cdots a_1)^{-1}(a_{n-1}\cdots a_1)$ before $a_n$ in the RHS of above the equation, and get that the RHS is equal to 
$$(a_1\cdots a_{n-1})(a_{n-1}\cdots a_1)^{-1}(a_{n-1}\cdots a_1)a_n (a_{n-1}\cdots a_1)^{-1} a_n^{-1}.$$
We are almost done: the last four terms form a commutator, whereas the first two terms form a long commutator with $n-1$ symbols, which by induction, should be in $G'$. Thus, $X\subseteq G'$ (Derek Holt concludes this by a simpler argument in comment after question). Obviously $X$ contains $G'$.
A: Based on the suggestion by Derek Holt, here´s my answer.
Let be $x \in X$, then $x = a_1\cdots a_na_1^{-1}\cdots a_n^{-1}$. Lets consider the rigth coset of $G'$ in $G$ with respect to $x$, $G'x \in G/G'$. We know that $G/G'$ is abelian, Thus
$$G'x = G'a_1\cdots a_na_1^{-1}\cdots a_n^{-1} = G'a_1a_1^{-1}\cdots a_na_n^{-1} = G'e = G'$$
Then we conclude 
$$G'x = G' \Longleftrightarrow x \in G'$$
Then $X \subseteq G'$, which completes the proof.
A: Here is my own proof. It has the advantage that it gives yet another characterization of the elements in the derived subgroup $K$.
Observe that every commutator $x$ can be written as $abc$ where $cba=1$ (easy exercise). Thus, every element $k$ of  $K$ can be written as a product $k=g_1...g_n$ with $g_n...g_1=1$. The inverse of $1$ being $1$, $k=1$ $k=$ what you want (just look at the inverse of the product being $1$).
A: Here's another proof.
Consider a long comutator $a_1 a_2 \dots a_n a_1^{-1} a_2^{-1} \dots a_n^{-1}$.
Since
$$
[a, b]ba = ab
$$
for all $a, b \in G$, we see that applying $[a_2, a_1]$ from the left we get
$$
[a_2, a_1]a_1 a_2 \dots a_n a_1^{-1} a_2^{-1} \dots a_n^{-1} = a_2 a_1 a_3 \dots a_n a_1^{-1} a_2^{-1} \dots a_n^{-1}.
$$
Applying $[a_3, a_2 a_1]$, we get
$$
[a_3, a_2 a_1][a_2, a_1]a_1 a_2 \dots a_n a_1^{-1} a_2^{-1} \dots a_n^{-1} = a_3 a_2 a_1 \dots a_n a_1^{-1} a_2^{-1} \dots a_n^{-1}.
$$
Eventually, this completely reverses the first half of the long commutator and we get the identity, so that its inverse is a product of commutators, hence it must be in $G'$ since $G'$ is a group.
