How to prove that $(AB)^t = B^tA^t$ The proof given in my book (and I came up with as well) is:

However, the part that throws me off is line #3 where they do $\Sigma A_{jk} B_{ki} = \Sigma B_{ki} A_{jk}$
I understand that multiplication of real numbers is commutative and you can do the last step. However, you could've just as easily left the expression without swapping the two and arrived at the expression: $(AB)^t = A^tB^t$, right?
What am I missing here?
 A: Wrong. The $ij$ entry of the matrix $A^tB^t$ is $\sum_kA_{kj}B_{ik}$, not $\sum_kA_{jk}B_{ki}$.
A: You are not missing much, just the fact that by convention we usually write
$$(AB)_{ij}=\sum_k A_{ik}B_{kj}$$
and not (although equivalent)
$$(AB)_{ij}=\sum_k B_{kj}A_{ik}$$
because in the first case “ikkj becoms ij”
A: What they want to make clear in line 3 is, that when you want to calculate the entry $ij$ of the transposed matrix, you can sum over the $i$-th column of $B$ and $j$-th row of $A$. So this is the same when you sum over the $i$-th row of $B^T$ and the $j$-th column of $A^T$. And that's what they do in the next line.
Again, this is the calculation:
$$(AB)^T_{ij}=(AB)_{ji}=\sum_{k=1}^n A_{jk} B_{ki}=\sum_{k=1}^n B_{ki}A_{jk}=\sum_{k=1}^n B^T_{ik}A^T_{kj}=(B^TA^T)_{ij}$$
A: What you have is that the $(i,j)$-th element of $C^t$ is
$$ \sum_k A_{jk}B_{ki} $$
where $A_{jk}$ is the $(j,k)$-th element of $A$ and $B_{ki}$ is the $(k,i)$-th element of $B$
Then because of the commutativity of the multiplication of the underlying field, you get that $A_{jk}B_{ki} = B_{ki}A_{jk}$, since both $A_{jk}$ and $B_{ki}$ are just elements of your field. Then you get in total:
$$ \begin{split}
\sum_k A_{jk}B_{ki} &= A_{j1}B_{1i} + A_{j2}B_{2i} + \dots \\
&= B_{1i} A_{j1} + B_{2i} A_{j2} + \dots \\
&= \sum_k B_{ki} A_{jk}
 \end{split}$$
EDIT:

you could've just as easily left the expression without swapping the
  two and arrived at the expression: $(AB)^t = A^tB^t$, right?

No, the textbooks wants to show that $(AB)^t = B^t A^t$. Because generally speaking $(AB)^t \neq A^t B^t$. Example:
$$\left(
\begin{bmatrix}
    1 & 0 \\
    0 & 0 
\end{bmatrix}
\begin{bmatrix}
    0 & 1 \\
    0 & 0 
\end{bmatrix}
\right)^t
=
\begin{bmatrix}
    0 & 1 \\
    0 & 0 
\end{bmatrix}
^t
=
\begin{bmatrix}
    0 & 0 \\
    1 & 0 
\end{bmatrix}
$$
But
$$
\begin{bmatrix}
    1 & 0 \\
    0 & 0 
\end{bmatrix}^t
\begin{bmatrix}
    0 & 1 \\
    0 & 0 
\end{bmatrix}^t
=
\begin{bmatrix}
    1 & 0 \\
    0 & 0 
\end{bmatrix}
\begin{bmatrix}
    0 & 0 \\
    1 & 0 
\end{bmatrix}
=
\begin{bmatrix}
    0 & 0 \\
    0 & 0 
\end{bmatrix}
$$
And as José Carlos Santos already pointed out, $\sum_k A_{jk}B_{ki}$ is not the $(i,j)$-th entry of $A^tB^t$, but as your textbooks shows, it is the $(i,j)$-th entry of $B^tA^t$
