Obviously! One could have noticed that we are not going to prove the actual Fermat's Last Theorem, rather we will prove a slight variation of the original statement (just by adding a extra condition). This condition changes the difficulty of the problem by day and night.
Statement: Let \(x, y, z \in \mathbb{N}\), \(n > 2\) and \(n \ge z\), then \(x^n + y^n = z^n\) has no non-trivial integer solutions.
Proof. For the sake of contradiction, suppose \(x^n + y^n = z^n\) has a non-trivial integer solution. Without the loss of generality, assume \(x < y\). So, we have \[ \begin{aligned} x^n = z^n - y^n = (z - y)(z^{n-1} + yz^{n-2} + \cdots + y^{n-1}) \end{aligned} \] Since, we have \(x < z\) and \(y < z\), by substituting \(y\) we have \[ \begin{aligned} (z - y)(z^{n-1} + yz^{n-2} + \cdots + y^{n-1}) &> (z - y)(y^{n-1} + yy^{n-2} + \cdots + y^{n-1}) \\ &> (z - y)ny^{n-1} \\ &> ny^{n-1} \end{aligned} \] Also, since \(n \ge z\), we have \(x^n > ny^{n-1} > zy^{n-1} > zx^{n-1} > x^n\), a contradiction. Therefore, there exists no such pair of positive integers satisfying \(x^n + y^n = z^n\). Hence proved