# Verify proof: $a^{n}-b^{n} = (a-b) \sum\limits_{k=0}^{n-1} a^{k}b^{n-1-k}$

A short disclaimer: I do know this question has been asked multiple times here and several answers (including combinatorics) have been given already. However, among all these posts, I did not find anywhere the answer by induction (as I have it understood) and hence, I thought to ask here for verification.

One should prove the following: $$a^{n}-b^{n} = (a-b) \sum\limits_{k=0}^{n-1} a^{k}b^{n-1-k} \quad (\star)$$

Proof:

$$1^{\text{st}}$$ Step: For $$n=1$$ it holds.

Induction Step: We assume that $$(\star)$$ is true for $$n$$. In order to show, that it is also valid for $$n+1$$, we calculate

$$a^{n+1}-b^{n+1}=a^n\cdot a-b^n\cdot b=a^n\cdot a-b^n\cdot a+b^n\cdot a-b^n\cdot b$$

$$=(a^n-b^n)a+b^n(a-b)\qquad \text{at this point we use } (\star)$$

$$=a(a-b)(a^{n-1}+ba^{n-2}+b^2a^{n-3}+\dots+b^{n-2}a+b^{n-1})+b^n(a-b)$$

$$=(a-b)(a^{n}+ba^{n-1}+b^2a^{n-2}+\dots+b^{n-2}a^2+b^{n-1}a+b^n)$$

Does this suffice as a proof? Or, is there any mistake in the induction step that I don't see right now?

• "Does this suffice as a proof? " No, you didn't say where you exactly used the induction hypothesis (although it is clear of course). Also the induction hypothesis is written with $\sum_k$, whereas you apply it without writing a sum symbol (which again is not important, of course, but just a bit inconsistent). Commented Nov 18, 2022 at 12:57
• It's fine, although I don't see why you need induction: you can just expand the right-hand side of $(*)$ and cancel most of the terms to show that it equals the left-hand side. Commented Nov 18, 2022 at 13:06