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:
$a^5-a^4b+ a^4b-a^3b^2+a^3b^2-a^2b^3+a^2b^3-a^1b^4+a^1b^4-b^5 = a^5 - b^5$
 A: \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 
$$b^{i-1}\le a^{i-1}$$
$$a^{n-i}b^{i-1}\le a^{n-1}$$
$$\sum_{i=1}^na^{n-i}b^{i-1}\le \sum_{i=1}^na^{n-1}=na^{n-1}$$
A: For the equalty,
By telescoping: We have,
$$(a-b)\sum_{i=1}^na^{n-i}b^{i-1}=a\sum_{i=1}^na^{n-i}b^{i-1}-b\sum_{i=1}^na^{n-i}b^{i-1},$$
which can be rewritten
$$\sum_{i=1}^na^{n+1-i}b^{i-1}-\sum_{i=1}^na^{n-i}b^i=a^n+\sum_{i=2}^na^{n+1-i}b^{i-1}-b^n-\sum_{i=1}^{n-1}a^{n-i}b^i,$$
which on simplifying gives
$$a^n-b^n+\sum_{i=1}^{n-1}a^{n-i}b^i-\sum_{i=1}^{n-1}a^{n-i}b^i=a^n-b^n.$$
By induction: You wish to prove
$$a^n-b^n=(a-b)\sum_{i=1}^n a^{n-i}b^{i-1}.$$
Let's try induction. For the $n=1$ case we have
$$(a-b)a^0b^0=a^1-b^1,$$
so the base case holds.
Now suppose the general case is true and consider the $n+1$ case. We have
$$(a-b)\sum_{i=1}^{n+1}a^{n+1-i}b^{i-1}=(a-b)\sum_{i=1}^n a^{n+1-i}b^{i-1}+(a-b)a^0b^n,$$
which can be rewritten
$$a(a^n-b^n)+(a-b)b^n=a^{n+1}-ab^n+ab^n-b^{n+1}=a^{n+1}-b^{n+1},$$
so in fact the general case holds.

For the inequality, you can use induction too:
$$a^n-b^n\leq (a-b)na^{n-1}.\tag{*}$$
The base case clearly holds since $a^1-b^1\leq (a-b)a^0$. Now suppose (*) holds and consider the $n+1$ case,
$$a^{n+1}-b^{n+1}=a^na-b^nb=(a^n-b^n)(a+b)-a^nb+b^na\tag{1}$$
But since $a>b$, then
$$(1)\leq (a^n-b^n)(a+b)-a^nb+a^nb=(a^n-b^n)(a+b)-b(a^n-b^n),$$
which we can write
$$(a^n-b^n)a\leq a(a-b)na^{n-1}=(a-b)na^n\leq(a-b)(n+1)a^n,$$
as required.

A: Geometric progression
\begin{align}
(a-b)\sum_{i=1}^{n}a^{n-i}b^{i-1}&=(a-b)a^{n-1}\left(\frac{1-\left(\frac{b}{a}\right)^n}{1-\frac{b}{a}}\right)\\
&=a^n-b^n.
\end{align}
 Can you complete the answer now?
A: Hint: Use that $$\sum_{i=1}^na^{n-i}b^{i-1}=\frac{a^n-b^n}{a-b}$$ if $$a\ne b$$
