Fundamental theorem of arithmetic1: every whole number can be split into a product of prime numbers in a unique way.
Examples:
$$6 = 2 \cdot 3 = (2,3)$$
$$8 = 2 \cdot 2 \cdot 2 =2^3 = (2, 2, 2)$$
What is prime factorization of one? An empty sequence, because it is the multiplicative identity and an empty product is the identity.
Now given this theorem, let's think of 6 from a different POV:
$$6 = (2,3)$$
Then dividing by a prime p is equivalent to dropping a term of value p from the sequence on the right hand side.
Now let's take a bold step and declare 0 has a factorization. Name it $$F$$(actorization).
$$0 = F$$
But $$\frac{0}{2} = 0$$, so dropping a term from $$F$$ should give us the factorization of zero. Which is $$F$$ itself!
This dropping property holds for any prime $$p$$, and by extension any nonzero number at all2.
So $$F$$:
- Has to be infinite length, else dropping a term from it would strictly shorten it.
- Contains every prime infinitely many times since 0 can be divided arbitrarily many times without changing.
So the prime factorization of 0 is all the prime numbers, repeated a infinite number of times. Unlike every other standard natural number, not a finite sequence.
$$0 = (2^\infty,3^\infty,5^\infty...) = \prod_p p^\infty$$
So if you take every prime and multiply them all by each other infinitely many times, you get zero. Trippy.