Proof of inequality $b^n-a^n<(b-a)nb^{n-1}$ when $00$. 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:
$$
b^n-a^n < (b-a)nb^{n-1}, \ \text{where} \ 0<a<b.
$$
I would really like to prove this inequality myself, and I tried rewriting the RHS. This produced
$$
b^n-a^n<n(b^n-a\cdot b^{n-1}).
$$
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.

note that b^n - a^n = (b-a)(b^n-1 + a*b^n-2 + a^2*b^n-3 + ... + b*a^n-2 + a^n-1) ...(1)
also since for any k>0 x^k is an increasing function , and hence a < b implies a^k < b^k
therefore , a^i * b ^(n-i-1) < or = b^i  * b^(n-i-1) = b^n-1 for i = 0,1,2,...,n-1 .... (2). 
So from (1) and (2) we have b^n - a^n < or =(b-a)nb^n-1 which is what we were required to prove.
