The fundamental theorem of arithmetic states that every integer greater than 1 can be represented uniquely as a product of prime numbers. We will first show the existence of these product and then we will prove the uniqueness.
Given an integer \(n \gt 1\), \(n\) prime or composite. If \(n\) is composite, then there should exist a prime that divides it, and then the resulting number would either be a prime or composite. If we keep repeating this process, we will end up with \(n=p_1p_2p_3 \ldots p_k\) where each \(p_i\) is a prime and where some primes may occur multiple times. This shows that some product for \(n\) exists.
It could be possible that \(n=p_1p_2p_3 \ldots p_k = q_1q_2q_3 \ldots q_h\), where \(h \ge k\) and where \(q_i\) is either a prime or 1. If \(n=p_1p_2p_3 \ldots p_k = q_1q_2q_3 \ldots q_h\), then:
Since \(p_2p_3 \ldots p_k\) is an integer, then \(p_1 | q_1q_2q_3 \ldots q_h\). According to this lemma, either \(p_1 | q_1\) or \(p_1 | q_2q_3 \ldots q_h\). If \(p_1 | q_2q_3 \ldots q_h\), then either \(p_1 | q_2\) or \(p_1 | q_3 \ldots q_h\). By repeating this process, we can claim that one of these has to be true: \(p_1 | q_1\), \(p_1 | q_2\), \(p_1 | q_3\), ..., \(p_1 | q_{h-1}\) or \(p_1 | q_h\). Let's assume that \(q_1\) is the prime that can be divided by \(p_1\). Since \(q_1\) is a prime, then \(p_1=q_1\), therefore \(\frac{q_1}{p_1}=1\). So now we have:
Using the same argument as above, we can claim that one of \(q_i\) should be divisible by \(p_2\). Let's assume \(q_2\) is the one that is divisble by \(p_2\). Then we have:
If we keep repeating the same argument as above, eventually we would end up with:
This means only \(k\) numbers of \(q_i\) integers where primes, and all those \(q_i\) primes were equal to some \(p_j\) primes. This proves the uniquesness.