Fundamental Theorem of Arithmetic says every positive number has a unique prime factorisation. Question: If 1 is neither prime nor composite, then how does it fit into this theorem?
Let us remember that an empty product is always 1. Hence, 1 has the empty product as its prime factorization. This product is vacuously a unique product of primes.
It has (uniquely!) zero prime factors.
I think you have simply misinterpreted the theorem. It should be stated as "...every positive number greater than one has a unique prime factor." .c.f. http://en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic
The OP hasn't misinterpreted the theorem. Every nonzero integer can be written as a product of primes.(GTM84 P.3) Just the exponents are all zeros...
You need to change the theorem because anything that works for this contradicts the theorem. Any natural number greater than 1 can be written as a product of prime factors.
You could say that i is prime because it's only factors are $i$ and $1$. It has no prime factors on the non-complex list of numbers therefore it is prime correct? $i^4=1$ (nothing in the theorem bans complex numbers) so you can say that $i^4$ is it's prime factorization. Being empty could be a solution but i^4 is also a solution meaning there are $2$. But there can't be. Also $i^8$ and $i^12$ work meaning there are infinite. So that means that $i$ is not prime but then what are it's other factors? I think that $1=i^4$ is better than $1= . $You could argue that you cannot count $i^4$ because $i^8$ also works but if $1= $ then $12= *2^2*3$ which contradicts the theorem more than $i^4$. So, what is the prime factorization of $1$?