Square Root of Prime is Irrational/Proof 1

Proof
Let $p$ be prime.

that $\sqrt p$ is rational.

Then there exist natural numbers $m$ and $n$ such that:

Any prime in the prime factorizations of $n^2$ and $m^2$ must occur an even number of times because they are squares.

Thus, $p$ must occur in the prime factorization of $n^2 p$ an odd number of times.

Therefore, $p$ occurs as a factor of $m^2$ an odd number of times, a contradiction.

So $\sqrt p$ must be irrational.