Different proofs of $\,a^n-b^n =(a-b)\sum_{i=0}^{n-1} a^i b^{n-1-i} $?

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.

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}