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I was given the problem:

For finite sets $A_1, A_2,\dotsc , A_n$ define their Cartesian product $\prod_{i=1}^n A_i$ as the set of all $n$-sequences $(x_1, x_2,\dotsc, x_n)$, where $x_i \in A_i$ for every $i = 1, 2, \dotsc, n$. Find a formula expressing the cardinality of $\prod_{i=1}^n A_i$ in terms of cardinalities $|A_1|, |A_2|,\dotsc , |A_n|$.

And I am struggling to understand what it is actually asking for, could someone explain it to me please, thanks. :)

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    Do you know what the Cartesian product is? For instance, if $A_1 = \{1,2\}$ and $A_2 = \{3,4,5\}$ could you explicitly write down $\prod_{1 \leq i \leq 2} A_i$? – 2017-01-19
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    @MeesdeVries This is all prossible combinations right? So for that example {(1,3), (2, 3), (1, 4) ....}, But I have never seen the notation: ∏ 1≤i≤n Ai before, what does that mean? – 2017-01-19

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$\prod_{1\le i\le n}A_i$ is the cartesian product, that is, all finite sequences $(a_1,\ldots,a_n)$ such that $a_i \in A_i$ for each $i=1,\ldots,n$. How many such sequences can you choose? $|A_1|$ choices for $a_1$, ..., $|A_n|$ choices for $a_n$. Therefore $$ \left|\prod_{1\le i\le n}A_i\right|=\prod_{1\le i\le n}|A_i|. $$

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    So is this question asking for a formula for the carnality of the cartesian product of $A_1, A_2,\dotsc , A_n$ ? – 2017-01-19
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    Yes. ____________ – 2017-01-19
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    So $\prod_{1\le i\le n}|A_i| $= $|A_1| * |A_2| * \dotsc * |A_n|$ – 2017-01-19
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    Again, yes :) _____ – 2017-01-19
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    @PaoloLeonetti Do you know if this hold for infinite sized sets as well? I suppose that would require a definition of multiplication of 'infinities'? Is there such a definition? – 2017-01-19
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    @Bram28 Assuming the axiom of choice, multiplication of infinite cardinal numbers is not difficult. If either $\kappa$ or $\mu$ is infinite and both are non-zero, then $\kappa \cdot \mu= \max\{\kappa, \mu\}$. In particular, $\kappa^n=\kappa$ for all $n\ge 1$. See here: https://en.wikipedia.org/wiki/Cardinal_number – 2017-01-19
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    @PaoloLeonetti Thanks!! If you don't mind another question ... The Axiom of Choice is about sets, right? So how does multiplication come in? Do we somehow combine the Axiom of choice with Peano Arithmetic? Or will multiplication be defined as a set-theoretic operation? Is there an accessible proof (or informal explanation) that derives $|A| \times |B| = max\{ |A|,|B|\}$ for the infinite and non-zero case? ... If this has no easy answer, let me know, and I'll post this as a question ... Thanks! – 2017-01-19
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    @Bram28 Well, you can find the answer in every introduction about cardinal arithmetics. Take a look here http://www.math.uchicago.edu/~may/VIGRE/VIGRE2009/REUPapers/Murphy.pdf and here http://euclid.colorado.edu/~monkd/m6730/gradsets06.pdf – 2017-01-19
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    @PaoloLeonetti Great, thanks for the links!!! ... Oh wow, just looked at them ... So for cardinal numbers, we simply *define* $|A| \times |B| = |A \times B|$. That's ... Interesting! Thanks again! – 2017-01-19
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    @Bram28 You're welcome ;) – 2017-01-19
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    @PaoloLeonetti +1 for the extra help! I learned something new :) – 2017-01-19
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We know that $$|A\times B|=|A|\times|B|\qquad(1)$$.

We want to show $$\left|\prod_{i=1}^nA_i\right|=\prod_{i=1}^n|A_i|$$ is true for any natural number $n$, where $\prod_{i=1}^n|A_i|=|A_1|\times|A_2|\times\dotsc\times|A_n|$. So, we have use induction.

The base case $n=1$ ($|A_1| = |A_1|$) is trivial. Now suppose inductively that $\left|\prod_{i=1}^nA_i\right|=\prod_{i=1}^n|A_i|$. We want to show $$\left|\prod_{i=1}^{n+1}A_i\right|=\prod_{i=1}^{n+1}|A_i|.$$ Now we need to show $$\left|\prod_{i=1}^{n+1}A_i\right|=\left|\left(\prod_{i=1}^nA_i\right)\times A_{n+1}\right|\qquad(2),$$ i.e., the cardinality of the set $\prod_{i=1}^{n+1}A_i$ is equal to the cardinality of the set $\left(\prod_{i=1}^nA_i\right)\times A_{n+1}$. So $$\begin{aligned}\left|\prod_{i=1}^{n+1}A_i\right|&=&\left|\left(\prod_{i=1}^nA_i\right)\times A_{n+1}\right|&\qquad\text{by }(2)\\&=&\left|\prod_{i=1}^nA_i\right|\times |A_{n+1}|&\qquad\text{by }(1)\\&=&|A_1|\times|A_2|\times\dotsc\times|A_n|\times|A_{n+1}|&\qquad\text{by induction hypothesis}\\&=&\prod_{i=1}^{n+1}|A_i|.\end{aligned}$$

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    This is exact, but I think the OP was asked a formula, but not asked to prove it. – 2017-01-19
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    @CristianGz Do you know if (1) also holds for infinities? If so, how am I to think about multiplying 'infinities'? Is there a definition for that? – 2017-01-19
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    @JeanMarie: Agree. I deleted the answer. But I though it can be useful. – 2017-01-19
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    @Bram28, the statement is true to infinite sets. It is possible using bijective functions to state equal cardinality between two sets. Check https://en.m.wikipedia.org/wiki/Cardinality – 2017-01-19