I need some feedback on this proof I wrote that:
$$\forall n\in\mathbb{N} \text{ assumed the existence of a factorization of } n \text{ as } n = p_1p_2\cdots p_k, \text{ where } p_i, (1 \leq i \leq k) \text{ is a prime number, such factorization is unique.}$$
My proof:
Due to the well-ordering theorem, if there exists a set of natural numbers that can be factorized in two different ways, there exists a least element of such set.
Let $n$ be the least number that has two different prime factorizations. Then $$n = p_1p_1\cdots p_k$$ and also $$n = q_1q_2\cdots q_s$$ Assuming all common factors have been crossed out, $\forall 1 \leq i, j \leq s,k, p_i \neq q_j$.
Without loss of generality, assume $q_1 > p_1$. Now let $m:= (q_1-p_1)q_2\cdots q_k$ . Then $$m = q_1q_2\cdots q_k - p_1q_2\cdots q_k = n - p_1q_2\cdots q_k \tag{ $\dagger$ }$$ Then $m<n$.
Plugging the two factorizations of $n$ into our expression $(\dagger)$, we get:
(1) $m = p_1p_2\cdots p_s-p_1 q_2\cdots q_k$
(2) $m =q_1q_2\cdots q_k - p_1q_2\cdots q_k$
Note that the factor $q_1$ is present in expression (2), but not in (1). Therefore, those two expressions are different prime factorizations of $m$. However, we have that $m<n$ and what we found contradicts the assumption that $n$ was the least element to have such a property.
Therefore, no natural number has such a property. □
Well, I'd consider a simpler proof by noting that if a prime $p$ divides a product $ab$, then $p$ divides a factor, i.e. $a$ or $b$.
Then in your ansatz, $p_1\cdots p_k = n =q_1\ldots q_l$, the prime $p_1$ divides the rhs and must be equal to a prime $q_j$. W.l.o.g. take $j=1$.
Then one can apply the shortening rule such that $p_2\cdots p_k=q_2\ldots q_l < n$. Now apply induction.