A bound on potential positive solutions of x^n + y^n = z^n

Suppose $(x,y,z)$ is a solution in $\mathbb{N}^+$ of the equation $x^n + y^n = z^n$, where $n > 2$. Show that $x,y > n$.

Note that $z > y$.

If $x^n + y^n = z^n$, then $x^n = z^n - y^n$ $= (z-y)(\sum_{k=0}^{n} {n \choose k} z^{n-k}y^k)$ $\geq \sum_{k=0}^{n} {n \choose k} z^{n-k}y^k$ $\geq {n \choose {n-1}} z y^{n-1}$ $= nzy^{n-1} > ny^{n-1}$. Thus $x^n > ny^{n-1}$. Likewise, $y^n > nx^{n-1}$.

Suppose, without loss of generality, that $x > y$. Then $y^n > nx^{n-1} > ny^{n-1}$, and so $y > n$. Thus $x > n$.