We prove existence and uniqueness separately.

Existence: We prove by strong induction on $n$ that every integer $n \geq 2$ can be expressed as a product of primes.

Base case: $n = 2$ is prime, so it is already a product of primes (a product with one factor).

Inductive step: Assume every integer $k$ with $2 \leq k < n$ can be expressed as a product of primes. Consider $n$.

If $n$ is prime, we are done. Otherwise, $n$ is composite, so $n = ab$ where $1 < a, b < n$. By the inductive hypothesis, both $a$ and $b$ can be expressed as products of primes: $a = p_1 p_2 \cdots p_r, \quad b = q_1 q_2 \cdots q_s$

Therefore: $n = ab = p_1 p_2 \cdots p_r q_1 q_2 \cdots q_s$

This is a product of primes, completing the induction.

Uniqueness: Suppose $n$ has two prime factorizations: $n = p_1 p_2 \cdots p_r = q_1 q_2 \cdots q_s$ where each $p_i$ and $q_j$ is prime (not necessarily distinct, and ordered arbitrarily).

Since $p_1$ divides the left side, it divides the right side: $p_1 \mid q_1 q_2 \cdots q_s$. By Euclid's Lemma, $p_1$ divides some $q_j$. Without loss of generality, reorder so that $p_1 \mid q_1$.

Since both $p_1$ and $q_1$ are prime and $p_1 \mid q_1$, we must have $p_1 = q_1$.

Cancel $p_1$ from both sides: $p_2 p_3 \cdots p_r = q_2 q_3 \cdots q_s$

Repeat this process. Each time, we match a prime from the left with a prime from the right and cancel. If $r < s$, we eventually get: $1 = q_{r+1} q_{r+2} \cdots q_s$

This is impossible since each $q_i > 1$. Similarly, $s < r$ leads to a contradiction. Therefore, $r = s$, and after reordering, $p_i = q_i$ for all $i$. The factorization is unique.