The proof of $\sqrt{2}$ is not rational number via fundamental theorem of arithmetic. I assume that $\sqrt{2}$ is positive number satisfies $(\sqrt{2})^2=2$. 

proof) Let $m$, $n$ as natural number,$\ $ $M$ is the number of prime factor of $m$,$\ $ $N$ is also the number of prime factor of $n$. For example, $m=12=2^2\cdot3$, $M$ is $3$. 

Then, if $\sqrt{2}$ were rational number, it could be expressed as a fraction $\frac{m}{n}$ in lowest terms. 

If $\sqrt{2}=\frac{m}{n}$, $m^2=2\cdot n^2$. Then, $m^2$ has $2M$ prime factor, $n^2$ has $2N$ prime factor. LHS has even prime factor, RHS has odd prime factor.  This is a contradiction to fundamental theorem of arithmetic.  

Therefore the initial assumption—that $\sqrt{2}$ can be expressed as a fraction—must be false.

Is there any problems? If not, I think this proof is simple and easier than ordinary proof -contradiction to the property of lowest terms.

p.s. This proof may apply to any root of $n$-th power $\sqrt[n]{a}$,  $n\in\mathbb{N}$ iff $a$ is prime.
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.
The step Bill Dubuque refers to can be made more explicit with this argument:  

If P(x) is the prime factorization of x then by definition P(x)P(x) is equal to xx. The combination of the sets P(x) and P(x) [allowing duplication of elements] is when multiplied out, by FTA, the prime factorization of only one number, which because it is P(x)P(x) can only be xx. Consequently if M(P(x)) is the size of P(x), M(P(xx)) = M(P(x)) + M(P(x)).