Which was defined first to represent $\underbrace{a+a+a+\cdots+a+a+a}_{n \text{ terms}}$? $n\times a$ or $a \times n$? When we are talking about multiplication, we often use it without knowing which one was defined first and which one was defined because of its commutative property.
Here I want to know which one was defined first?
$$\underbrace{a+a+a+\cdots+a+a+a}_{n \text{ terms}} \equiv n \times a $$
or
$$\underbrace{a+a+a+\cdots+a+a+a}_{n \text{ terms}} \equiv a \times n $$
?
 A: The fact that multiplication is commutative makes the question not have any particularly meaningful answer - we can define it either way and derive that the other way is correct as well. In fact, we could, given addition, define the multiplication of integers by the properties:
$$(a_1+a_2)(b_1+b_2)=a_1b_1+a_2b_1+a_1b_2+a_2b_2$$
$$1\cdot 1=1.$$
$$1\cdot -1=-1\cdot 1=-1.$$
Which is entirely symmetric and can quickly be seen to uniquely define the operation of multiplication. Obviously, given the symmetry of the definition, this makes multiplication commutative - but moreover, since we can supply an entirely symmetric definition, it makes the question of which definition was "first" meaningless, since we can define things in which neither is first, and neither is implicit in being first.
That aside, perhaps your intention is not purely about the values that multiplication takes or its algebraic properties, but rather something more general. Like, if you wish to be consistent across hyperoperators, note that:
$$a^b = \underbrace{a\cdot\ldots\cdot a}_{b\text{ times}}$$
so, perhaps the first definition you give is better in that since, as the left operand controls the value in the sum and the right one controls how many sums are taken. (Historically, I'll bet that both definitions showed up independently, and the fact that they were equivalent was understood before the modern notation of multiplication showed up)
A: While this doesn't directly address the 'which came first' question, I think it's worth pointing out an implicit assumption in the original question:
Multiplication is commutative over natural numbers, but it's not over ordinals; there $a\cdot b$ has a specific meaning, and roughly corresponds to $b$ copies of $a$ 'laid head to toes' — in other words, $a+a+a+\cdots+a$.  In particular, $\omega\cdot 2\equiv \omega+\omega$ is two copies of $\omega$ (i.e., the natural numbers) with an order such that every element of the 'second' copy of $\omega$ is greater than ever element of the first; it looks like $\langle 0_0, 1_0, 2_0, \ldots, 0_1, 1_1, 2_1, \ldots\rangle$.  On the other hand, $2\cdot\omega\equiv 2+2+2+\cdots$ is $\omega$ copies of $2$ next to each other - and this is precisely the order $\omega$ itself ($2\cdot\omega$ is $\langle0_0, 1_0, 0_1, 1_1, 0_2, 1_2, \ldots\rangle$, and it's easy to find an order-preserving mapping from this to $\langle 0, 1, 2, 3, 4, 5, \ldots\rangle$).
A: Something multiplied by a number... that something does not have to be a number, it  can be a length, an area, an event 
I would think that "$a$ multiplied by $m$, or $a$ taken $m$ times, or $a$ $m$ times,  these should be written as $a \times m$. 
However $m$ times $a$  should be written as $m \times a$. 
The answer may also depend on the language, tradition, point of view.
Interesting question.
A: It seemed to me there would have been a method used by the Romans to perform multiplication. Their method might as well be taken as "first". So I searched and found the following. The order of 'a x n' in practice is the smaller number first.
See http://www.professorfekete.com/articles%5CCh-2%20HOW%20%20DID%20%20THE%20%20ROMANS%20%20DO%20%20MULTIPLICATION.pdf

"For example, you want to find 58 × 249. Write the two numbers next to 
one another; then double one and halve the other. (It is preferable, though not 
mandatory, to halve the smaller and double the larger.) If in halving there is a 
remainder, ignore it. Repeat the process until the number in the halving column 
is reduced to 1. Then in the doubling column cross out every number that stands 
opposite to an even number in the halving column, and sum up the rest. The sum 
is exactly the required product: "
58 249
29 498 
14 996
7 1992 
3 3984 
1 7968 
sum the right side (not crossed out)
14442 = 58 × 249 
