Geometric Series Proof, I can' get other terms to cancel out! 
When I calculate the answer out I get $x^n +x^{n-2}+ x^3-1$. What am doing wrong? This version of the proof is linked the question $a^n-1$ being divisible by $a-1$, which is why I need this version of the proof.
 A: \begin{eqnarray}
(1-x) \sum_{k=0}^{n-1} x^k &=& \sum_{k=0}^{n-1} x^k - \sum_{k=0}^{n-1} x^{k+1} \\
&=& \sum_{k=0}^{n-1} x^k - \sum_{k=1}^{n} x^{k} \\
&=& 1 + \sum_{k=1}^{n-1} x^k - \sum_{k=1}^{n-1} x^{k} - x^{n} \\
&=& 1-x^n
\end{eqnarray}
A: Hint:
$\color{blue}{x^n}+x^{n-1}+\ldots + x^3+x^2+x$
$-\left( x^{n-1}+\ldots + x^3+x^2+x+\color{blue}1\right)$
Writing $\color{blue}1$ outside the bracket:
$\color{blue}{x^n}+\left(x^{n-1}+\ldots + x^3+x^2+x\right)$
$-\left( x^{n-1}+\ldots + x^3+x^2+x\right)-\color{blue}1$
A: You have forgotten to fully include the effect of the terms hidden inside the $\cdots$. When you do this you will see that there is a $x^{n-2}$ term in the first $(\cdots)$ and a $x^3$ term inside the second $(\cdots)$ which should cancel with the ones you got in your final answer.
To see this more clearly we can write your next to final expression as
$$(x^n + x^{n-1} + \cdots+x^3+x^2+x) - (x^{n-1} + x^{n-2} + \cdots +x^2 +x +1)\\=\text{(all powers of $x$ from $x^1$ up to $x^n$)} - \text{(all powers of $x$ from $x^0$ up to $x^{n-1}$)}$$
Any power $x^k$ where $k$ is less than $n$ and larger than $0$ is included in both of two terms above so it gets subtracted away
$$(\cdots + x^k + \cdots) - (\cdots + x^k + \cdots) = (\cdots +  0 + \cdots)$$
Since this is the case for all $k=1,2,3,\ldots,n-1$ we are only left with two terms. The first term $x^n$ in the (all powers of $x$ from $x^1$ up to $x^n$) and the last term $x^0 = 1$ in (all powers of $x$ from $x^0$ up to $x^{n-1}$). This gives us the result $x^n-1$.
