How to prove that, for each $n\in \mathbb N$ $$
\frac{x^{n+1}-1}{x-1} = 1 + x + x^2 + \dots + x^n
$$
where $x\neq 1$, $x\in \mathbb R$.
I am really tired to prove that questions. I can not understand any one. Please help me.....How to prove that, for each $n\in \mathbb N$ (using Mathematical induction)
My try 
how to complete this steps? 

This is equivalent to $(x-1)(1+x+\cdots x^n) = x^{n+1}-1$
Base case, $n=0$,
$$(x-1)(x^0)=x^{0+1}-1$$
Now assume it is true up to $n-1$, so
$$(x-1)(1+x+\cdots x^{n-1}+x^n) = [(x-1)(1+x+\cdots x^{n-1})] + (x-1)(x^n) $$
$$= [x^n - 1] + [x^{n+1}-x^n] =x^{n+1}-1 $$
Now divide the $(x-1)$ out to RHS
$$(1+x+\cdots x^{n-1}+x^n) = \frac{x^{n+1}-1}{x-1}$$
which we can do so long as $x\neq 1$.
