Problem: If $x$,$y$,$z$ and $n>1$ are natural numbers with $$x^n+y^n = z^n$$ then show that x,y and z are all greater then $n$.

My approach, from Fermat's Theorem we know that $x^n + y^n = z^n$ does not hold for $n>2$. And since there doesn't exist any triplet $x,y,z \in \{1,2\}$, $x,y,z$ will always be greater than $n$.

Here is the approach given in the book,
$$z>x,y$$
Suppose,$$y\ge x$$
Since, $z>y$, we have $z\ge y+1$. Now
$$ x^n = z^n - y^n$$
$$=(z-y)(z^{n-1}+z^{n-2}+...y^{n-1})$$
$$>((y+1)-y)nx^{n-1}$$
(This above step is what I do not understand)
then,
$$x^n>nx^{n-1}$$or$$ x>n$$

What I do not understand is the third step and from where did the $nx^{n-1}$ come? Also is my proof not sufficient.