I know that since $a>1$ is composite, then it can be broken down into a product of prime factors, by Fundamental Theorem of Arithmetic. So then $a=p_1p_2\dots p_k$ for some natural number k. Then, I notice that since $a=p_1p_2\dots p_k$, then there is a prime factor $p$ in that product of primes that divides $a$, therefore $p$ divides $a$.
But how do I show that $p$ is less than or equal to $\sqrt{a}$?
If $p$ is greater than $\sqrt{a}$ then $a/p\le\sqrt{a}$. (Otherwise, both $a/p$ and $p$ are greater than $\sqrt{a}$, so $a=(a/p)p>(\sqrt{a})^2$ which is absurd.) Then consider any prime factor of $a/p$.
$a=p_1p_2\dots p_k$, with $k\ge2$, so, if you let $p=p_1$ be the smallest prime that divide $a$, you have $a=ps$ with $s\ge p$.
$$ \sqrt{a}=\sqrt{ps}\ge\sqrt{p^2}=p $$
Since $a$is composite it has proper divisors that are larger than 1. Let $p$ be the smallest such divisor. It is prime. It is not hard to argue that $p^2 \le n$.
