$S=1+x+x^2+x^3+...x^{n-1}=\dfrac{x^n-1}{x-1}$
Now I wanted to prove it using induction. Using induction for $n=1$, gives us $S=1$ which is prime. OK!
Now assume for $n=m$, it is true.
Then $\dfrac{x^n-1}{x-1}$ is prime $\implies m=prime$
Now if $m\ne2$ or $m\ne1$, then $\notin m+1$ such that $m+1$ is $ \space prime$
Thus w/o loss of generality, let $m+k$ where $k\in \mathbb{N}$ and $k<m$ such that $m+k$ is prime and also assume $m+k$ is the next prime after $m$
Now required to prove $\dfrac{x^{m+k}-1}{x-1}$ is prime
We see here that $\dfrac{x^{m+k}-1}{x-1}=1+x+x^2+.....x^{m+k-1}=1+x+x^2+...x^m+x^{m+1}+...x^{m+k-1}$
Now we know that $\underbrace{1+x+x^2+...x^{m-1}}_{\text{is prime by hypothesis}}+x^{m}+...x^{m+k-1}=\mathbb{P}+x^m(1+x+x^2+...x^{k-1})$
Now as $k<m$, we can say that $(1+x+x^2+...x^{k-1})$ is also prime if we use strong induction.
So now we get $\mathbb{P}+\underbrace{x^m\underbrace{(1+x+x^2+...x^{k-1})}_{\text{is prime by using strong induction}}}_{\text{is composite}}\implies \mathbb{P}+\text{a composite number}$ which is a possible candidate for prime.
But how can I say that if I add it, I must get a prime again?
Or Is the approach entirely wrong?Please suggest with reasoning.
Ok so by popular demand let us use contradiction here
Assume $n$ is not prime.
Then what to do next?