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? 

Assume it holds for $n-1$. then $1+x+x^2\dots+x^{n-1}=\dfrac{x^{n}-1}{x-1}$. From this we get:
$1+x+x^2\dots+x^{n-1}+x^n=\dfrac{x^{n}-1}{x-1}+x^n=\dfrac{x^n-1+x^n(x-1)}{x-1}=\dfrac{x^n-1+x^{n+1}-x^n}{x-1}=\dfrac{x^{n+1}-1}{x-1}$
Meaning it holds for $n$
