number theory - Using the fundamental theorem of arithmetic to prove $sqrt{text{prime}}notinBbb Q$

I need to use the fundamental theorem of arithmetic to show that

if $p$ is prime then $\sqrt p$ is irrational.

So far I've stated that $\sqrt p=m/n$ where $m,n$ are positive integers, then $pn^2=m^2$. Now I can factor $m$ and $n$ into primes but I don't know where to go on from there.

Answer

Given a prime $p$ and some $n\in\mathbb{N}^*$, let we define:

$$\nu_p(n)=\max\{k\in\mathbb{N}: p^k\mid n\}.\tag{1}$$
Since $\mathbb{Z}$ is a UFD we have $\nu_p(ab)=\nu_p(a)+\nu_p(b)$. In particular, $\nu_p$ of a square is always an even number. If we assume that $\sqrt{p}=\frac{m}{n}$ with $m,n\in\mathbb{N}^*$, we also have
$$p n^2 = m^2. \tag{2}$$

However, such identity cannot hold, since $\nu_p(\text{RHS})$ is an even number and $\nu_p(\text{LHS})$ is an odd number. It follows that $\sqrt{p}\not\in\mathbb{Q}$ as wanted.