Commutator Subgroup of Direct Product equals the Direct Product of the Commutator Subgroups I need to prove that $(G_{1}\times G_{2})^{\prime} = G_{1}^{\prime} \times G_{2}^{\prime}$, where $G^{\prime}$ denotes the commutator subgroup of $G$ - i.e., $G^{\prime}=[G,G]$, the subgroup generated by all commutators of elements of $G$. 
Recall that the commutator of two elements $x$ and $y$ in a group $G$, denoted $[x,y]$ is defined to equal $x^{-1}y^{-1}xy$.
I have posted my proof as an answer below. Could somebody please take a look at it and let me know if it's okay? If not, please let me know what I need to do in order to fix it. Thank you. :)

Edit: I have been informed that my answer given below actually shows only equality of the set of commutators, and that $G^{\prime}$ consists of products of commutators. So, this question is no longer a proof check, but I am asking specifically how to fix what I have written below in order to make it actually answer the thing I set out to prove. I am having a bit of trouble understanding some of the hints given me thus far, and am going to need more detail in any answers given in order for them to be helpful.
 A: Let $(g_{1},g_{2})\,,\,(h_{1},h_{2}) \in G_{1}\times G_{2}$. Then, $\begin{align}[(g_{1},g_{2})(h_{1},h_{2})]=(g_{1},g_{2})^{-1}(h_{1},h_{2})^{-1}(g_{1},g_{2})(h_{1},h_{2}) \\ =(g_{1}^{-1},g_{2}^{-1})(h_{1}^{-1},h_{2}^{-1})(g_{1},g_{2})(h_{1},h_{2}) = (g_{1}^{-1}h_{1}^{-1}g_{1}h_{1},g_{2}^{-1}h_{2}^{-1}g_{2}h_{2}) \\ = ([g_{1},h_{1}],[g_{2},h_{2}]) = ([g_{1},h_{1}],e_{G_{2}})(e_{G_{1}},[g_{2},h_{2}])\end{align}$.
So, each commutator in $G_{1} \times G_{2}$ is the product of one in $G_{1}$ with one in $G_{2}$.
Therefore, $\mathbf{(G_{1}\times G_{2})^{\prime}\subseteq G_{1}^{\prime}\times G_{2}^{\prime}}$.
Now, to get the inclusion in the other direction, let $(g,h) \in G_{1}^{\prime} \times G_{2}^{\prime}$. Then, $g = [g_{1},g_{2}]\in G_{1}^{\prime}$ and $h = [h_{1},h_{2}] \in G_{2}^{\prime}$.
So, 
$\begin{align}(g,h) = ([g_{1},g_{2}],[h_{1},h_{2}]) = (g_{1}^{-1}g_{2}^{-1}g_{1}g_{2}, h_{1}^{-1}h_{2}^{-1}h_{1}h_{2}) \\ = (g_{1}^{-1},h_{1}^{-1})(g_{2}^{-1},h_{2}^{-1})(g_{1},h_{1})(g_{2},h_{2}) \\ = (g_{1},h_{1})^{-1}(g_{2},h_{2})^{-1}(g_{1},h_{1})(g_{2},h_{2}) \\= [(g_{1},h_{1}),(g_{2},h_{2})] \end{align} $ 
where $(g_{1},h_{1}), (g_{2},h_{2}) \in G_{1} \times G_{2}$. 
Therefore, the product of a commutator in $G_{1}$ with a commutator in $G_{2}$ is a commutator in $G_{1} \times G_{2}$. Thus, $\mathbf{G_{1}^{\prime} \times G_{2}^{\prime} \subseteq (G_{1} \times G_{2})^{\prime}}$.
And so, we get our equality, $G_{1}^{\prime} \times G_{2}^{\prime} = (G_{1} \times G_{2})^{\prime}$.
