Any composite number can be expressed as a unique product of prime factors.
My attempt was:
Let $b \in \mathbb{Z}$ and is a composite number. One of the ways to find the divisors of a number is to divide it by an integer $a$, such that $\frac{b}{a} =p$ with $p \in \mathbb{Z}$. Also if I divide this quotient again, I will get another divisor of $b$, the proof of this is:
If $(1)\frac{b}{a} = p$, i'll prove $(2)\frac{p}{d} = c, \{c,d\} \in \mathbb{Z}$, then $d$ is divisor of $b$.
From $(1) b = ap$ and from $(2) d = \frac{p}{c}$, therefore $\frac{b}{d} = \frac{apc}{p} = ac$, then is divisor.
With this proved i can say that if $\frac{p}{d} = c \to p =dc = \frac{b}{a}$
Notice that $p = dc$, with $d,c \neq 1$ only happens when $p$ isn't a prime number, per example $12/3 = 4$, and $4$ is a divisor of $12$, now divide the quotient by other integer to get other divisor of $12$, i.e $4/2 = 2$, therefore $2$ is other divisor of $12$, but from here $2 = 2\cdot 1$ and i can't get new divisors. And here is my trick, since if you never get that $p = dc$ is a prime number then you could continue dividing and obtaining divisors infinitely and therefore would be an infinite number.
Well, this only proves that, by successively dividing a composite number, you will arrive at a prime number, otherwise you would have infinite divisors.
Therefore it would be enough to express any composite number as a product of any of its divisors and divide it recursively until it reaches prime numbers (as already shown). Per example, $1920 = 24 \cdot 80$. Lets divide the divisor $24$ that is $24/6 = 4 \to 6 \cdot 4$, lets divide the divisor $6$, that is $6/2 = 3 \to 6 = 2\cdot 3$, lets divide $4$ that is $4/2 = 2 \to 4 = 2 \cdot 2$, therefore $24 = 6\cdot 4 =(2 \cdot 3) \cdot (2 \cdot 2)$ and this is what I mean "recursively".
Then, is this non-formal proof valid for the fundamental theorem of arithmetic? or at least correct? Have you a improvement to this proof?
If a number is composite, it has at least one non-trivial factor (i.e. not one nor the number itself). If this factor is prime, we are done with it; otherwise, we can decompose it similarly. By doing so, we recursively decompose the number in prime factors.
But we have to prove that this process is finite. It indeed is because a proper factor is necessarily smaller than the number and we will eventually be stopped before $1$.
For uniqueness, assume that there are two distinct decompositions. Strike out the common factors. Then taking the simplified decompositions modulo some of the primes, on one side we get $0$, and on the other a nonzero product.