How is an ordered pair that belongs to $R^2$ x $R^3$ different from a list that belongs to $R^5$? According to this link (at 3:05), the author says that $R^2$ x $R^3$ ≠ $R^5$ as the object of $R^2$ x $R^3$ [(($x_1$,$x_2$),($x_3$,$x_4$,$x_5$)) : $x_1$,$x_2$,$x_3$,$x_4$,$x_5$ ∈ R] is different from an object of $R^5$ [($x_1$,$x_2$,$x_3$,$x_4$,$x_5$) : $x_1$,$x_2$,$x_3$,$x_4$,$x_5$ ∈ R] in the sense that the objects in $R^2$ x $R^3$ is an ordered pair, whereas the objects in $R^5$ are lists of length five.

My doubt here is, how is an ordered pair that belongs to $R^2$ x $R^3$ different from a list that belongs to $R^5$ ? 
 A: They are only formally different. The length of an object in $\mathbb R^2×\mathbb R^3$ is two (an ordered pair may be considered a length-$2$ list), while that of an object in $\mathbb R^5$ is five.
The first-glance intuition that they are the same is formalised by a trivial bijection between the two spaces.
A: There is a sense, in which you can consider them exactly equal. For simplicity consider $R\times R^2$ and $R^3$.
We define an $n$-tuple as follows:

$(x_1,x_2)$ is the set $\{x_1,\{x_1,x_2\}\}$.
By induction, we define an $n$-tuple as $(x_1,\dots,x_n)$ as
  $(x_1,(x_2,\cdots,x_{n-1}))$.

It is now easy to see that $R\times R^2=R^3$.
To illustrate, note that $$(x_1,x_2,x_3)=(x_1,(x_2,x_3))=\{x_1,\{x_1,(x_2,x_3)\}\}=\{x_1,\{x_1,\{x_2,\{x_2,x_3\}\}\}\}.$$ So $R^3$ consists of all such sets.
Now consider $R\times R^2$. An element of $R\times R^2$ is of the form $(a,b)$ with $a\in R$ and $b=(b_1,b_2)\in R^2$. In other words an arbitrary element of $R\times R^2$ is
$$(a,b)=\{a,\{a,b\}\}=\{a,\{a,(b_1,b_2)\}\}=\{a,\{a,\{b_1,\{b_1,b_2\}\}\}\}.$$
So $R\times R^2$ consists of all such sets.
At this point it is clear that $R\times R^2=R^3$. Similarly, $R^n\times R^m=R^{mn}$.
