# Proof of inequality $b^n-a^n<(b-a)nb^{n-1}$ when $0<a<b$ and $n>0$.

I am working through some properties of $$\mathbb{R}$$ and I stumbled upon the following theorem:

Theorem 1.21: For every real $$x>0$$ and every integer $$n>0$$, there is one and only one positive real $$y$$ such that $$y^n=x$$.

The proof of this theorem, as stated in Rudin's Principles of Mathematical Analysis heavily relies on the following inequality: $$b^n-a^n < (b-a)nb^{n-1}, \ \text{where} \ 0

I would really like to prove this inequality myself, and I tried rewriting the RHS. This produced $$b^n-a^n Clearly, the value given inside the parentheses on the RHS will be positive, as $$b>a$$, but I fail to see how this gives me the inequality itself. Does it maybe have to do something with the Archimedean Property of $$\mathbb{R}$$?

Any help would be much appreciated.

• Hint: factor $b^n-a^n$ as the product of $b-a$ and something else. May 20, 2020 at 13:21
• $(b^n-a^n)=(b-a)(b^{n-1}+b^{n-2}a+\ldots + a^{n-1})$ May 20, 2020 at 13:22