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So far, I have convinced myself that the statement

$$b^{n+1} - a^{n+1} = (b^n + ab^{n-1} + ... + ba^{n-1} + a^n)(b-a)$$

is true. I verified for $n=1,2,3$.

My first guess for proving this is to use induction, so (roughly speaking), I need to prove that

$$b^{k+2} - a^{k+2} = (b^{k+1} + ab^{k} + ... + ba^{k} + a^{k+1})(b-a)$$

However, I am not sure how to show this last statement. Is induction the right approach in this case? Does anyone have a hint for me?

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    See https://en.m.wikipedia.org/wiki/Geometric_progression – 2017-02-04
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    The question has multiple downvotes, yet the tag is algebra-precalculus and some, albeit minimal, effort is shown as well as correct formatting. Not everyone comes to this site with an advanced background. I regularly see questions on this site that are blatant ploys to get homework answered -- with no effort -- and users jump to post full answers – 2017-02-04
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    Just to clarify for others, this isn't a homework question. This equality is used in a real analysis proof that I am studying, and I wanted to learn how to prove it. – 2017-02-04

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A direct approach is possible. (Induction is implicit in applying the distributive law to general finite sums.)

$$(b-a) \sum_{k=0}^n b^{n-k} a^k = b \left(b^n + \sum_{k=1}^n b^{n-k} a^k \right) - a \left(a^n + \sum_{k=0}^{n-1} b^{n-k} a^k \right) \\ = b^{n+1} - a^{n+1} + \sum_{k=1}^n b^{n-k+1} a^k - \sum_{k=0}^{n-1} b^{n-k} a^{k+1}. $$

Change the index in the first sum to $j = k-1$ and you should be able to finish on your own.

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    Thank you for your response. This approach worked like a charm! Could you elaborate on how the inductive step is implicit in applying the distributive law? – 2017-02-04
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    @EternusVia: You're welcome. I point that out because you asked for a "proof" and we take some algebraic properties for granted when manipulating sums in this way. Given the distributive law $a(c_1 + c_2) = ac_1 + ac_2$ can we generalize to $a \sum_{k=1}^n c_k = \sum_{k=1}^n (ac_k)$? It is intuitively obvious but technically it should be proved if we only assume that the basic law is a given. By induction if $a \sum_{k=1}^{n-1} c_k = \sum_{k=1}^{n-1} (ac_k)$ is true then $a \sum_{k=1}^{n} c_k = a (\sum_{k=1}^{n-1}c_k + c_n) = a\sum_{k=1}^{n-1} + ac_n = \ldots$ follows. – 2017-02-04