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? 

First of all you verify that this hold for some number, i.e. $n=0$.
Then you assume that it holds for some $n=k$.
And then you need to prove the result for $n=k+1$ by using that assumption.
Now, $\frac{x^{k+1}-1}{x-1}=x^k+\frac{x^k-1}{x-1}$. From the Induction Hypothesis for $n=k$ we have that $\frac{x^{k}-1}{x-1}=1+x+x^2+...+x^{k-1}$. 
Hence, $\frac{x^{k+1}-1}{x-1}=x^k+\frac{x^k-1}{x-1}=x^k+(1+x+x^2+...+x^{k-1})$. So it holds for $n=k+1$ also and this complete the Induction.
Thus $\frac{x^{n+1}-1}{x-1}=1+x+x^2+...+x^n$ holds for any $n$.
