On the other hand, you can do the same by noting that $\sqrt[n]{2}$ is a root of
$$x^n - 2 = 0$$
and rational root theorem says that the only rational roots, if any, must be one of
$$\{-1, 1, -2, 2\}$$
none of which are proper candidates for $n > 1$, so $\sqrt[n]{2}$ is not rational.
Edit : Further, you can just use Gauss's lemma by noting that
$$x^n - 2$$
is irreducible over $\mathbb{Z}[x]$, as the sign changes occur in the interval $(2, 1)$ and the negative counterpart assuming for all $n > 1$, none of which contains integers, and thus by Gauss's lemma is also irreducible over $\mathbb{Q}[x]$.
I will try to give you a VERY quick overview on the strategy of the proof of FLT. Of course I cannot avoid to use very technical tools, such Galois representation. If you don't know about these stuff, I hope you can at least follow the "shape" of the argument.
The starting observation is the following: for $n\in\mathbb N$, let FLT($n$) be the statement "there are no triples $(a,b,c)$ of integers with $abc\neq 0$ such that $a^n+b^n=c^n$". Then it is elementary to see that FLT($d$) implies FLT($n$) whenever $d\mid n$. Therefore it is sufficient to prove FLT($p$) for every odd prime $p$ and for $n=4$ in order to prove FLT($n$) for all positive integers $n\geq 3$. The cases $p=3$ and $n=4$ were already known to Euler, I believe, and can be proven by elementary methods, so we can assume that $p\geq 5$.
Now the non-elementary math comes in. Suppose that there is a triple of integers $(a,b,c)$ which contradicts FLT($p$) for some prime $p$. One can construct the following elliptic curve over $\mathbb Q$:
$$E_{a,b,c}\colon y^2=x(x-a^p)(x-b^p)$$
It is possible to show (assuming wlog that $(a,b,c)$ is a coprime triple, $a\equiv -1\bmod 4$ and $2\mid b$) that this is a semistable elliptic curve whose conductor is $\prod_{l\mid abc}l$.
Moreover, Serre and Frey proved the following theorem: let
$$\overline{\rho}_{a,b,c}\colon \text{Gal}(\overline{\mathbb Q}/\mathbb Q)\to GL_2(\mathbb F_p)$$
be the residual Galois representation at $p$ attached to $E_{a,b,c}$. Then:
- $\overline{\rho}_{a,b,c}$ is absolutely irreducible;
- $\overline{\rho}_{a,b,c}$ is odd;
- $\overline{\rho}_{a,b,c}$ is unramified outside $2p$ and flat at $p$
The key idea is then the following: suppose that we can show that for every (semistable) elliptic curve $E$ over $\mathbb Q$ the representation $\overline{\rho}_E$ is the residual representation of the $p$-adic Galois representation attached to a weight $2$ newform $f_E$. Then we can apply a theorem of Ribet, using the properties of $\overline{\rho}_{a,b,c}$, to show that for $\overline{\rho}_{a,b,c}$ we can choose such a newform in $S_2(\Gamma_0(2))$. But then we have a big problem: this space is $0$-dimensional! Therefore we are led to a contradiction, and there cannot be any triple $(a,b,c)$ contradicting FLT($p$).
This whole argument was already known long before Wiles' proof. But the thing that was missing was the proof of the (nowaday) so-called modularity theorem: every elliptic curve over $\mathbb Q$ is modular, i.e. its $p$-adic Galois representation is the $p$-adic Galois representation attached to a weight $2$ newform.
Even though this was the only missing part, it is by far the most difficult one! Wiles was able to prove it for semistable curves, which was enough for proving FLT, but some years later the result has been improved to all elliptic curves over $\mathbb Q$ by Breuil, Conrad, Diamon and Taylor.
Anyway, if you are interested in more details, you can read the (amazing) first chapter of "Modular forms and Fermat's last theorem", by Cornell, Silverman and Stevens eds. Of course it is a math textbook, so you need some background in these type of topics in order to understand it.
Best Answer
Many experts suspect that Wiles's proof of FLT can be converted into a proof in Peano arithmetic which is much weaker than set theory, see Are there non-standard counterexamples to the Fermat Last Theorem? and What is known about the relationship between Fermat's last theorem and Peano Arithmetic?
If this is possible then we only need axioms of set theory sufficient to derive Peano arithmetic. This would mean that axioms of choice, infinity and replacement are not needed, they are used to deal with infinite sets. What is left is a theory of finite sets, which is more or less equivalent to arithmetic, see Systems the Peano axioms can be derived in. But so far whether it can be done or not is an open question. As Colin McLarty writes in What Does It Take to Prove Fermat's Last Theorem:
Generally speaking, tracking which axioms are actually used in a proof is a tedious but routine exercise. It is not very informative though. Most proofs are using results whose use can be avoided to simplify or shorten proofs. Wiles's proof uses the aformentioned modularity thesis, "universes", axiom of choice constructions, and other very strong tools that are not essential. As is it uses all ZFC axioms. Calculus, as is, uses them all as well (possibly excepting replacement), but most of it can be done without the axiom of choice.
On the other hand, figuring out what is minimally needed to prove a complex theorem is a non-trivial question in its own right, and is related to the so-called independence results, a field in mathematical logic.