Suppose $A_1, \dots , A_n$ is a list of finite
sets.
Then

\[ \textstyle \num{\prod_{i = 1}^{n}} A_i = \prod_{i = 1}^{n} \num{A_i} \]

As before, the proof will be by induction.
Future editions will include.

We highlight one simple consequence: if $\num{A}$
is finite, then $\num{A^n} = \num{A}^n$.
Suppose $X_1, \dots , X_k$ is a list of
nonempty sets and $f_2, \dots , f_k$ are
functions mapping $X_1, X_1 \times X_2, \dots ,
X_1 \times \cdots X_{k-1}$ to $\powerset{X_2},
\powerset{X_3}, \cdots, \powerset{X_k}$ such that
there exists natural numbers $n_1$, $n_2$, …,
$n_k$,

\[ \begin{aligned} n_1 &= \num{X_1} \\ n_2 &= \num{f(x_1)} \quad \text{for all } x_1 \in X_1 \\ n_3 &= \num{f(x_1, x_2)} = n_3 \quad \text{for all } (x_1, x_2) \in X_1 \times f(x_1) \\ n_4 &= \num{f(x_1, x_2, x_3)} = n_3 \quad \text{for all } (x_1, x_2, x_3) \in X_1 \times f_2(x_1) \times f_3(x_1, x_2) \\ &\vdots&\\ n_k &= \num{f(x_1, x_2, \dots , x_{k-1})} \quad \\& \quad\quad \text{for all } (x_1, x_2, \dots , x_{k-1}) \in X_1 \times f_2(x_1) \times \cdots \times f_k(x_1, \dots , x_{k-1}) \\ \end{aligned} \]

Then\[ \num{\Set{x: \set{1, \dots , k} \to \cup_{i = 1}^{k}}{x_1 \in X_1, x_2 \in f_2(x_1), \dots , x_k \in f_k(x_1, \dots , x_{k-1}}} \]

is $\prod_{i = 1}^{k} n_i$ Future editions will include the proof.

The English is much clearer.
Suppose we have to do $k$ *tasks* (here,
the term task is left undefined).
The first task can be done in $n_1$ ways, and
after it has been completed, no matter how, the
second task can be done in $n_2$ ways.
Further, after the first two tasks have been
completed, in whatever manner, there are $n_3$
ways of doing the third task.
And so on.
Then the number of ways to complete all tasks
is $n_1 n_2 \cdots n_k$.
This conclusion, often taken as a principle, or
axiom, is called the fundamental
principle of counting.

Consider the usual 52 playing cards cards. How many ways are there to stack them as a deck? First the bottom card, we have 52 choices. For the next to bottom card, we have 51 choices—we can pick any card except the bottom one. For the third to bottom card, we have 50 choices—and so on. By applying the above proposition, we can deduce that there are $52\cdot 51\cdot 50\cdots1$ ways of arranging the deck.

Denote the set of cards by $C$. For the above proposition, $k = 52$ and $X_1, \dots , X_k$ are so that $X_i = C$ for all $i = 1, \dots , 52$. The function $f_2(c_1) = C - \set{c_1}$. The function $f_3(c_1, c_2) = C - \set{c_1, c_2}$. And so on, so that the function $f_k(c_1, \dots , c_{k-1}) = C - \set{c_1, \dots , c_{k-1}}$. Notice that given distinct arguments to these functions, the size of the sets $f_1(c_1)$ is always 51. Likewise, the size of $f_2(c_1, c_2)$ is 50, $f_3(c_1, c_2, c_3)$ is $49$ and so on, so that the size of $f_k(c_1, \dots , c_k)$ is $1$, regardless of the choice of $c_i$, $i = 1, \dots , 51$.