proving the inequality $n \cdot a^{n-1} \cdot(b-a) < b^n - a^n < n \cdot b^{n-1} \cdot (b-a)$ let $b>a>0$ and $n>1$ I need to prove that:
$n \cdot a^{n-1} \cdot(b-a) < b^n - a^n < n \cdot b^{n-1} \cdot (b-a)$
I though of proving it using induction, or to build a function on where n is the variable but failed both times.
 A: There are two ways to prove this.
Method 1:
Using the fact that $\dfrac{b^n-a^n}{b-a}=b^{n-1}+a^{n-2}b+\cdots+a^{n-1}$, then
$$b^{n-1}+a^{n-2}b+\cdots+a^{n-1}>a^{n-1}+a^{n-1}+\cdots+a^{n-1}=na^{n-1}\\b^{n-1}+a^{n-2}b+\cdots+a^{n-1}<b^{n-1}+b^{n-1}+\cdots+b^{n-1}=nb^{n-1} \\na^{n-1}<\dfrac{b^n-a^n}{b-a}<nb^{n-1} \\ na^{n-1}(b-a)<b^n-a^n<nb^{n-1}(b-a)$$
Method 2:
Let $f(x)=x^n$, then $f'(x)=nx^{n-1}$.
By Mean Value Theorem, we know that $f'(a)<\dfrac{f(b)-f(a)}{b-a}<f'(b)$.
$$na^{n-1}<\dfrac{b^n-a^n}{b-a}<nb^{n-1} \\ na^{n-1}(b-a)<b^n-a^n<nb^{n-1}(b-a)$$
A: Let $x={b\over a}$, so that ${b^n-a^n\over b-a}=b^{n-1} {1-x^n\over 1-x}=b^{n-1}(1+x+...+x^{n-1})$. 
As $0\leq x<1$, we deduce   $b^{n-1} \leq {b^n-a^n\over b-a} \leq n b^{n-1}$, and the result as $a<b$
A: I would try induction. Try using the fact that
\begin{align*}
(b^{n-1}-a^{n-1})(b-a) = (b^n-a^n) - ab^{n-1} + ba^{n-1}
\end{align*}
For the first step of the induction (i.e. n=2), this helps to obtain the result. I didn't try the general induction step, but probably you can use the same statement again. 
A: $b^{n}-a^{n}=(b-a)\sum_{i=1}^{n}b^{n-i}\cdot a^{i-1}$
Clearly $b^{n-1}>b^{n-i}\cdot a^{i-1}>a^{n-1}$. Therefore,
$na^{n-1}\cdot (b-a)<b^{n}-a^{n}<nb^{n-1}\cdot(b-a)$
