mean value theorem question I was trying to solve the following:
Given: $0 < a < b$ and $n>1$,
prove:
$$na^{n-1}(b-a) < b^n-a^n < nb^{n-1}(b-a)$$
I managed to get this far using the mean value theorem:
$$a^n(b-a)<b^n -a^n<b^n (b-a)$$
Any idea how to continue?

Let $f(x)= x^n-a^n$ using the Mean Value Theorem
$$f'(c) = \frac{f(b)-f(a)}{b-a} = \frac{b^n - a^n}{b-a}$$ for some $c \in (a,b)$.
But $f'(c)= n \cdot c^{n-1}$ so 
$$\frac{b^n - a^n}{b-a} = n \cdot c^{n-1}$$
Now note that $n \cdot a^{n-1} < n \cdot c^{n-1} < n \cdot b^{n-1}$ so
$$ n \cdot a^{n-1} < \frac{b^n - a^n}{b-a} < n \cdot b^{n-1}$$
