The proof of $\sqrt{2}$ is not rational number via fundamental theorem of arithmetic.

Solution 1:

It looks OK, apart from one little typo:

$\sqrt{2}=\frac{n}{m}\implies 2m^2=n^2$, you got it the other way around.

Solution 2:

As to why it is easier than the usual proof: The usual proof doesn't use the fundamental theorem of arithmetic (or at least, can be avoided pretty easily), which actually takes some work to prove.

Using the fundamental theorem of arithmetic for $\mathbb{Q}$ (i.e: Any rational number can be uniquely written as a product of INTEGER powers of primes, this is an easy corollary of the statement for integers) you can prove a much more general claim:

If $r$ is a positive rational number and $n \in \mathbb{Z}$, then $r^{1/n}$ is rational if and only if the prime factorization of $r$ is of the form $$\prod_{p} p^{n e_p}$$ for some $e_p \in \mathbb{Z}$, with only finitely many $e_p \neq 0$.

In more fancy words: The multiplicative group of positive rationals is a free abelian group, where the primes form a basis.