A more rigorous way to prove this? I would like to prove the following statement
$$x^n-a^n=(x-a)\sum^{n-1}_{k=0}x^ka^{n-k-1},\qquad\forall n\in\Bbb N_0$$
I can easily prove it by induction using polynomial long division or series expansion however I am unsure whether or not these are "rigorous enough". I have thought about this for a while but is there a way to prove this statement in a more rigorous way? I think that the use of ellipses and "do this for the remaining $n$" gives me this uncertainty. Therefore I pose my question as follows:

Is there a rigorous way to prove this statement, or is the use of ellipses rigorous enough?

Thank you.
 A: Develop and change the index: (almost all the terms are canceled  except for $k=0$ and $k=n$):
$$(x-a)\sum^{n-1}_{k=0}x^ka^{n-1-k}=\sum^{n-1}_{k=0}x^{k+1}a^{n-1-k}-\sum^{n-1}_{k=0}x^ka^{n-k}=\sum^{n}_{k=1}x^{k}a^{n-k}-\sum^{n-1}_{k=0}x^ka^{n-k}\\=x^n-a^n$$
but I would say that all the proofs you mentioned are also rigorous. 
A: \begin{align}
& (x-a)\sum^{n-1}_{k=0}x^ka^{n-1-k} \\[10pt]
= & x \sum^{n-1}_{k=0}x^ka^{n-1-k} - a \sum^{n-1}_{k=0}x^ka^{n-1-k} \\[10pt]
= & \sum_{k=0}^{n-1} x^{k+1}a^{n-1-k} - \sum_{k=0}^{n-1} x^k a^{n-1-k+1} \\[10pt]
= & \sum_{\ell=1}^n x^\ell a^{n-\ell} - \sum_{k=0}^{n-1} x^k a^{n-k}.
\end{align}
Here we've let $\ell=k+1$.  But after that it does not matter whether the index that runs from $1$ to $n$ is called $\ell$ or $k$, so we have
\begin{align}
& \sum_{k=1}^n x^k a^{n-k} - \sum_{k=0}^{n-1} x^k a^{n-k} \\[10pt]
= & \left(\left( \sum_{k=1}^{n-1} x^k a^{n-k}\right)+ x^n a^0 \right) - \left( x^0 a^{n-0} + \sum_{k=1}^{n-1} x^k a^{n-k} \right) \\[10pt]
= & x^n - a^n
\end{align}
A: Yet another way:
\begin{eqnarray*}
(x-a)\sum_{k=0}^{n-1}x^{k}a^{n-k-1} &=&(x-a)a^{n-1}\sum_{k=0}^{n-1}(\frac{x}{%
a})^{k} \\
&=&(x-a)a^{n-1}\frac{1-(\frac{x}{a})^{n}}{1-\frac{x}{a}}=(x-a)\frac{%
a^{n}-x^{n}}{a-x}=x^{n}-a^{n}
\end{eqnarray*}
A: Using a series expansion is definitely rigorous enough. It's the most straightforward way to demonstrate the equality, using only basic algebra. 
