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

Proof  by induction on n:

1. For $n=1$ we get:  ${\displaystyle (X-a)\sum _{i=0}^{1-1}{a}^i{X}^{1-i-1}=(X-a)\cdot a^0{X}^0=X-a=X^1-a^1}$ .
    
2. Now suppose the equation to be valid for an arbitrary n. This implies for the next number $n+1$ :
    
   ${\displaystyle \begin{array}{ll}(X-a)\sum _{i=0}^n{a}^i{X}^{n-i}\hfill & =(X-a)(X\sum _{i=0}^{n-1}{a}^i{X}^{n-i-1}+a^n)\hfill \\ \hfill & =X(X^n-a^n)+a^n(X-a)\hfill \\ \hfill & =X^{n+1}-a^{n}X+a^{n}X-a^{n+1}\hfill \\ \hfill & =X^{n+1}-a^{n+1}\text{.}\hfill \end{array}}$