Show that for every $n\ge1$ and for every real number $x\ne 1$ $\dfrac {x^{n+1}-1}{x-1}=1+x+x^2+\cdots +x^n$ is valid.
By induction, put $n=1$ $x=2$ and you get for the base case:
$$\frac{x^2-1}{x-1}=x+1$$ $$x+1=x+1$$
Then consider $n=m$
$\dfrac {x^{m+1}-1}{x-1}=1+x+x^2+\cdots +x^m$ and assume that holds.
We then calculate $m=n+1$:
$\dfrac {x^{n+2}-1}{x-1}=1+x+x^2+\cdots +x^n+x^{n+1}$ I get:
$$\frac{x^{n+2}}{x-1}-\frac{1}{x-1}=1+x+x^2+\cdots+x^n+x^{n+1}$$
Multiply by $x-1$ and get:
$$x^{n+2}-1=(x-1)(1+x+x^2+\cdots+x^n+x^{n+1})$$
But how do we get further here?
When $n=1$ and $x=2$, the equation becomes $$\dfrac{2^2-1}{2-1}=\dfrac{3}{1}=3=1+2$$ so the statement holds true.
Full proof by induction:
We shall induct on $n$. In this proof, $x$ can be any real number except for $1$.
Base case: When $n=1$, we have $\dfrac{x^2-1}{x-1}=1+x$, which is clearly true.
Induction hypothesis: Suppose that when $n=k$, we have $\dfrac{x^{k+1}-1}{x-1}=1+x+x^2+\dots+x^k$.
Inductive step: Consider the case where $n=k+1$.We claim that $\dfrac{x^{k+2}-1}{x-1}=1+x+x^2+\dots+x^{k+1}$.
By the induction hypothesis, $$\begin{align} 1+x+x^2+\dots+x^{k+1} &=\dfrac{x^{k+1}-1}{x-1}+x^{k+1} \\ &=\dfrac{x^{k+1}-1}{x-1}+\dfrac{x^{k+2}-x^{k+1}}{x-1} \\ &=\boxed{\dfrac{x^{k+2}-1}{x-1}} \end{align}$$ This completes our induction.
Hence, for every $n\ge1$ and for every real number $x\ne 1$, $$\dfrac {x^{n+1}-1}{x-1}=1+x+x^2+\cdots +x^n$$
Alternative and more elegant proof:
Let $a=1+x+x^2+\dots+x^{n}$.
Then, $xa=x+x^2+\dots+x^{n}+x^{n+1}$.
Subtracting, we have $(x-1)a=x^{n+1}-1$.
Finally, dividing both sides by $x-1$, we get that $a=\boxed{\dfrac{x^{n+1}-1}{x-1}}$. Isn't that elegant?