# For every  $n\in ℕ$  we have:  $\left(X-a\right)\sum _{i=0}^{n-1}{a}^{i}{X}^{n-i-1}={X}^{n}-{a}^{n}$.

Proof  by induction on n:

1. For $n=1$ we get:  $\left(X-a\right)\sum _{i=0}^{1-1}{a}^{i}{X}^{1-i-1}=\left(X-a\right)\cdot {a}^{0}{X}^{0}=X-a={X}^{1}-{a}^{1}$ .

2. Now suppose the equation to be valid for an arbitrary n. This implies for the next number $n+1$ :

$\begin{array}{ll}\left(X-a\right)\sum _{i=0}^{n}{a}^{i}{X}^{n-i}\hfill & =\left(X-a\right)\left(X\sum _{i=0}^{n-1}{a}^{i}{X}^{n-i-1}+{a}^{n}\right)\hfill \\ \hfill & =X\left({X}^{n}-{a}^{n}\right)+{a}^{n}\left(X-a\right)\hfill \\ \hfill & ={X}^{n+1}-{a}^{n}X+{a}^{n}X-{a}^{n+1}\hfill \\ \hfill & ={X}^{n+1}-{a}^{n+1}\text{.}\hfill \end{array}$