TFA addresses this and states that's not the actual historical reason. If it was, you could keep 1 a prime and rephrase the theorem to "a unique product of primes greater than 1".

I don't follow what you mean by "the actual historical reason". The article clearly indicates that 1 went through a long period of not being commonly considered a prime (for bad reasons), then a brief period of being commonly considered a prime (for bad reasons), and is now once more not considered a prime (for good reasons). It calls out the reason why we switched to the current view as being, at the very least, extremely closely related to the reason you're claiming is not "the historical reason".

Your parent comment is correct (or so close that I think correcting them is a bigger mistake); that is the reason 1 is not prime. We don't consider 1 "not prime" as an accident of history, we have the knowledge now to make an informed decision. The "historical reason" has fallen by the wayside.

This is related, by the way, to the question of 0! (zero factorial). By definition, it's 1. The most obvious reason for that is that it fills a bunch of different holes in various formulae that use factorials (Taylor series, virtually any combinatoric formula...). But that's an extremely instrumentalist approach. Is there any reason why we might predict that 0! = 1 without knowing in advance from combinatorics that we'd really like it to be the case?

Sure. What happens if we multiply together the first zero positive integers? Obviously, we get the empty product: 1. Hence the definitions I learned, where a prime number has exactly one factor (not two), and 1 has zero, not enough to be prime.

Not sure what your point is.

The OP pretty much dismisses the article and cites the one thing that is well-known, less interesting, and not the focus of the article. The article is way more interesting and lists a variety or reasons (the one the OP quotes, historical reasons, wrong reasons, and even one final twist).

The OP's comment looks like a case of "did not RTFA". Color me unimpressed.

>It is not a prime because every integer greater than 1, must be made by a unique product of prime numbers.

That property is a result of the natural numbers, and a definition of primality that excludes 1. It is not true in general, in under a definition of primality that includes 1.

What about prime integers? Aren't they a product of 1 and themselves?

If you think of every integer as being a product of some other integers, e.g. the product of [a,b,c] then you could say the algorithm for that is: take the first number of the list, multiply it with the product of the rest of the numbers, when you get to an empty list, just apply the empty product, which is 1 for obvious reasons (1a is identity). So then [p] = p1. Then by this definition [] has no factors, it's the empty product. Of course you don't have to think of it like this, but it might be useful to do so.

This has been covered by anaphor, but to respond to the question directly:

No, a prime number, for example 5, is not the product of 1 and itself (at least, no more than it's the product of 1/3 and 15). It is the product of itself only, so it can be represented as a set of factors like {5}. 1 is the product of nothing, and would be represented as the set of factors {}.

Multiplication, obviously, just combines the factors of the multiplicands. 6 (= {2,3}) times 2 (= {2}) equals {2,2,3}, 12. Hopefully that makes it clear why 1 should be {}.

FTA:

What if we permitted 1 to be prime? In that case, 84 would also have the "prime" factorisation 1 x 1 x 1 x 2 x 2 x 3 x 7. That is, 84 could still be factorised, but it would no longer have a unique prime factorisation.

I did read the article. I was just pointing out that the poster's original statement seemed to exclude prime integers -- well, unless the empty product is a product of primes.