How to prove $a^n − b^n = (a − b) \sum_{i=1}^{n}a^{n-i} b^{i-1}\le (a − b)na^{n−1}$. Please tell if the problem can be solved using telescoping technique or not. 
If yes, how to prove $a^n − b^n = (a − b) \sum_{i=1}^{n}a^{n-i} b^{i-1}\le (a − b)na^{n−1}$ using that. It is given that $a,b \in \mathbb{R}{+},\, a\gt b,\, n \in \mathbb{N}.$ 
I tried as follows, but was unsuccessful to pursue:
$a^n − b^n = a^n+\sum_{i=1}^{n-1}(a^ib^{n-i}-a^ib^{n-i})-b^n=a^n+\sum_{i=1}^{n-1}a^ib^{n-i}-\sum_{i=1}^{n-1}a^ib^{n-i}-b^n$

Edit : based on the selected answer's comment.
Writing a few terms of the series, $\sum_{i=1}^n (a^{n+1-i}b^{i-1}-a^{n-i}b^i)$ get:
For $n =5$, get the terms as:
$i=1, \,\, a^{5+1-1}b^{1-1}-a^{5-1}b^1 = a^5-a^4b.$
$i=2, \,\, a^{5-1}b^{2-1}-a^{5-2}b^2 = a^4b-a^3b^2.$
$i=3, \,\, a^{5-2}b^{3-1}-a^{5-3}b^3 = a^3b^2-a^2b^3.$
$i=4, \,\, a^{5-3}b^{4-1}-a^{5-4}b^4 = a^2b^3-a^1b^4.$
$i=5, \,\, a^{5-4}b^{3-1}-a^{5-3}b^5 = a^1b^4-b^5.$
Adding all the terms, get:
$a^5-a^4b+ a^4b-a^3b^2+a^3b^2-a^2b^3+a^2b^3-a^1b^4+a^1b^4-b^5 = a^5 - b^5$

\begin{align}
(a-b)\sum_{i=1}^n (a^{n-i}b^{i-1}) &=\sum_{i=1}^n (a^{n+1-i}b^{i-1}-a^{n-i}b^i )\\
&=a^n+\sum_{i=2}^n a^{n+1-i}b^{i-1}-\sum_{i=1}^{n-1}a^{n-i}b^i - b^n \\
&= a^n+\sum_{i=1}^{n-1}a^{n-i}b^i-\sum_{i=1}^{n-1}a^{n-i}b^i-b^n\\
&=a^n-b^n
\end{align}
You might like to read the working backward to  be similar to what you attempted.
$b<a$, then 
$$b^{i-1}\le a^{i-1}$$
$$a^{n-i}b^{i-1}\le a^{n-1}$$
$$\sum_{i=1}^na^{n-i}b^{i-1}\le \sum_{i=1}^na^{n-1}=na^{n-1}$$
