UPDATE/WARNING: DO NOT READ (WASTE OF TIME)
The effort I put in here is now an embarrassment as it goes nowhere. I can't delete this posting since there is an answer, but if a moderator could delete it I would be much obliged.
I deleted all my work here (including comments and answers) and have rolled the question back to the original posting, except for this update.
Let $n$ be any integer greater than $1$.
Let $p$ be any prime factor of $n$.
Consider any factorization of $n$,
$\tag 1 n = ab$
Write both $a$ and $b$ using their $\text{Base-p}$ representations.
$\tag 2 a = a_0 + a_1 p + a_2 p^2 + a_3 p^3 + \dots$
$\tag 3 b = b_0 + b_1 p + b_2 p^2 + b_3 p^3 + \dots$
By the elementary theory found here, the only way $p$ can divide the product $ab$
$\quad$ = rhs of $\text{(2)}$ $\times$ rhs of $\text{(3)}$
is if $a_0 = 0$ or $b_0 = 0$.
Summarizing,
Proposition 1: Let $n$ be any integer with a prime factor $p$ and let $n = ab$.
Then $p$ must divide either $a$ or $b$.
Continuing, we know that if $n \gt 1$ the first integer greater than $1$ that divides into $n$, $\gamma(n)$, is a prime, the minimal prime factor.
Any integer $n$ can be factored in prime numbers, but an easy argument using proposition 1 guarantees that $\gamma(n)$ must appear in any factorization.
Theorem 2: Any two factorizations of an integer $n \gt 1$ into primes must agree.
Proof
Let
$\tag 4 n = \gamma(n)\, p_1 p_2 \dots$
$\tag 5 n = \gamma(n)\, q_1 q_2 \dots$
be any two factorizations. We can factor out $\gamma(n)$ in both representations. Now each remaining representation is either empty or contains more factors. If necessary, then, repeat, factoring out
$\gamma(\frac{n}{\gamma(n)})$
In this way all factors are matched up and uniqueness is guaranteed. $\quad \blacksquare$