Fundamental theorem of arithmetic^{1}: 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 all^{2}.

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.