Existence of a prime divisor

Statement

Let $n$ be a natural number with $n \ge 2$. Then there is a prime number $p$ such that $p \mid n$.

Remarks

One proof takes the smallest divisor $d$ of $n$ with $d > 1$. Each divisor $e$ of $d$ with $e > 1$ also divides $n$, so $e = d$. Thus $d$ is prime. This smallest divisor is the smallest prime factor of $n$. The result is also true for $n = 0$, since each prime divides $0$. It is false for $n = 1$.

Explanations

L1 human

Split $n$ into two smaller factors when you can, and repeat with one of the factors. The numbers get smaller each time, so the process stops at a number that you cannot split. This number is a prime, and it divides $n$. For example, $60 = 2 \cdot 30$, so $2$ is a prime divisor of $60$.