Let x,n ∈ $\mathbb{N}$ such that :
1+$x^1$+$x^2$+$x^3$+....+$x^{n-1}$ is prime .
Proof that n is prime .
I actually suppose that n isn't prime absurdly so n=p×k such as p is prime ,I also use the $a^n$-$b^n$ formula and find that $x^k$-1/$x^1$-1 so that means that k=1 and then n=p which is absurd. Does Anyone has another answer ?
If $x=1$, then clearly $n$ is prime. Let us consider $x>1$.
Suppose, towards a contradiction, that $n$ is composite. Then we may write $n=ab$ where $a,b\ge 2$. Thus
$$\begin{align*}1+x+\dots+x^{n-1}&=\frac{x^n-1}{x-1}\\ &=\frac{x^{ab}-1}{x-1}\\ &=\frac{(x^{b(a-1)}+x^{b(a-2)}+\dots+1)(x^b-1)}{(x-1)}\\ &=\frac{(x^{b(a-1)}+x^{b(a-2)}+\dots+1)(x^{b-1}+x^{b-2}\dots+1)(x-1)}{(x-1)}\\ &=(x^{b(a-1)}+x^{b(a-2)}+\dots+1)(x^{b-1}+x^{b-2}\dots+1)\end{align*}$$
which, provided $x>1$, is clearly composite. So we conclude that $n$ is in fact prime.
