By definition a prime has no factors, therefore it cannot be the PRODUCT of anything! Nothin times nothin = nothin. Don't believe everything you see on wiki, where anyone can be an editor for a day.
"Since any integer is a product of primes, there you have it . . . " My comment was not about the totient function, but the quoted statement. If it was true, then all integers would be composite, and we know that is false.
= http://en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic I note that the wikipedia article excludes 1, but a different article says 1 is a product of zero primes. This condition makes the Mobius function work.
A more formal source might say the product is over all primes with only a finite number of exponents at most being different from zero. \(n = \prod_{p \in \mathbb{P}} p^{\small n_p} \; ; \; \left{ p \in \mathbb{P} \, | \; n_p \neq 0 \; \right} \, \prec \, \mathbb{N}\) Alternately, a product over no more than a finite number of primes with the convention that the product of zero terms is 1, a convention that certainly helps in the partition of products. \(n = \prod_{k \in \left{ p \in \mathbb{P} \, | \; p \leq n \; \textrm{and} \; p|n \; \right}} k^{\small n_k}\) But yeah, 1 has a prime factor representation -- the empty representation. A theory of multiplication that didn't include 1 would be pretty weak.
