Sorry but I don't think it is nice. Because, in order to prove that ,you use Fermar's last theorem which is too strong. In fact, we can prove it very simple by number theory . For example:
Suppose it is rational then we can write it as
[imath] a^n= 2b^n[/imath] so [imath] a^n [/imath] must be even so
[imath] a [/imath] must be even .Put [imath] a=2k[/imath] then we get [imath] (2k)^n=2b^n[/imath]
[imath]2^{n-1}k^n=b^n [/imath]
so [imath]b [/imath] must be even.Then [imath]b=2k1 [/imath] .... And we can countinue it .So it is only ended when [imath] a=b=0[/imath] with is never happen. So [imath] \sqrt[n]{2} [/imath] can't be rational:D