Proof in Rudin, real numbers

Question

There is one thing in this proof I do not get. Why can he say that $k < y$?

Answer 1

Note that $$ k = \frac{y^n - x}{ny^{n-1}} = \frac 1n y - \frac{x}{ny^{n-1}} < \frac 1n y \leq y $$

Answer 2

You know that $$k=\frac{y^n-x}{ny^{n-1}}.$$

Then you can write it as: $$=\frac{y^n}{ny^{n-1}}-\frac{x}{ny^{n-1}}$$

Now, using $x>0$, $y>0$, $n>0$, the second term must be positive, thus you have the inequalities: $$<\frac{y^n}{ny^{n-1}}\leq\frac{y}{n}\leq y.$$

Using this chain of equalities/indequalities you have $k< y$.