Different proofs of $\,a^n-b^n =(a-b)\sum_{i=0}^{n-1} a^i b^{n-1-i} $? How many different proofs are there that
$a^n-b^n
=(a-b)\sum_{i=0}^{n-1} a^i b^{n-1-i}
$
for positive integer $n$
and real $a, b$?
You can use any techniques you want.
My proof just uses algebra,
summation, and induction,
but if you want to use invariant sheaves
over covalent topologies,
that is fine.
I decided that
I would try to produce
a proof by induction.
I find it interesting that
my proof shows that
if it is true for
$n-1$,
then it is true for $n+1$.
This means that two base cases
have to be proven:
$n=1$ and $n=2$.
Fortunately,
those are easy.
I am sure that my proof is known, 
but I do not recall having seen it before.
Here is the induction step:
$\begin{array}\\
a^{n+1}-b^{n+1}
&=a^{n+1}-a^nb+a^nb-b^{n+1}\\
&=a^{n+1}-a^nb +a^nb-ab^n +ab^n-b^{n+1}\\ 
&=(a-b)a^n +ab(a^{n-1}-b^{n-1}) +(a-b)b^n\\ 
&=(a-b)(a^n+b^n) +ab(a^{n-1}-b^{n-1}) \\ 
&=(a-b)(a^n+b^n) +ab((a-b)\sum_{i=0}^{n-2} a^i b^{n-2-i})
\ \  \text{(The induction hypothesis)} \\ 
&=(a-b)(a^n+b^n+ab\sum_{i=0}^{n-2} a^i b^{n-2-i}) \\ 
&=(a-b)(a^n+b^n+\sum_{i=0}^{n-2} a^{i+1} b^{n-1-i}) \\ 
&=(a-b)(a^n+b^n+\sum_{i=1}^{n-1} a^{i} b^{n-i}) \\ 
&=(a-b)\sum_{i=0}^{n} a^{i} b^{n-i} \\ 
\end{array}
$
 A: By telescopy $\ f_n = x^n\Rightarrow\  \displaystyle \overbrace{f_n-f_0 =\sum_{k\,=\,0}^{n-1}\left[f_{k+1}\:\!-\:\!f_k\right]}^{\textstyle x^{n} - 1 = \sum\ [x^{k+1}-x^k]_{\phantom{|_|}}}\, =\, (x\!-\!1)\sum_{k\,=\,0}^{n-1}\, x^{k} $
The sought result now follows by homogenization, i.e.  $\, x\to a/b\,$ then scaling by $\,b^n.$
Remark $\ $ The simple theorem employed to evaluate the above telescopic sum may be viewed as a discrete analog of the Fundamental Theorem of Integral Calculus
$$\begin{eqnarray} f &=& \sum \Delta f,\quad \Delta f(n) = f(n+1) -f(n)\\
f & =& \ \int D f,\quad D\, f(x) = f'(x)\end{eqnarray}$$
But the proof of the discrete analog is a trivial one-line induction exploiting telescopic cancellation.
A: Using
$$\sum_{k=0}^{n-1}x^k={1-x^n\over1-x}$$
we get
\begin{align}
a^n-b^n
&=a^n\left[1-\left({b \over a}\right)^n\right]\\
&=a^n\left[1-{b \over a}\right]\sum^{n-1}_{k=0}{b^k\over a^k}\\
&=\sum^{n-1}_{k=0}a^{n-k}b^k-\sum^{n-1}_{k=0}a^{n-k-1}b^{k+1}\\
&=(a-b)\sum^{n-1}_{k=0}a^{n-k-1}b^k
\end{align}
