Prove $n$ is prime. (Fermat's little theorem probably) 
Let $x$ and $n$ be positive integers such that $1+x+x^2\dots x^{n-1}$ is prime. Prove $n$ is prime

My attempt:
Say the above summation equal to $p$ $$1+x+x^2\dots x^{n-1}\equiv 0\text{(mod p)}\\
{x^n-1\over x-1}\equiv0\\
\implies x^n\equiv1\text{ (as $p$ can't divide $x-1$)}$$
How to proceed?
 A: This does not involve any Fermat's little theorem type tricks. Indeed, it follows from a very simple factorization trick. 
In some sense, if $n = pq$, then we arrange the $n$ powers $x^0,...,x^{pq-1}$ in a $p \times q$ box fashion, and then collect terms.
So, write $n = pq$ for some $1 < p,q < n$ if $n$ is not prime. Then:
$$
1 + ... +  x^{pq-1} = (1 + ... + x^{p-1}) + (x^{p} + ... + x^{2p-1}) + ... + (x^{p(q-1)} + ... + x^{pq - 1}) \\ = (1 + ... + x^{p-1}) (1 + x^p + x^{2p} + ... +  x^{p(q-1)})
$$
for example, if $n = 6 = 2 \times 3$ then
$$1 + ... + x^5 = (1+x)(1 + x^2 + x^4)$$, and if $n = 9 = 3 \times 3$ then $$1 + ... + x^8 = (1+x+x^2)(1+x^3+x^6)$$
A: Hint:
$$x^{ab}-1=(x^a-1)(x^{a(b-1)}+x^{a(b-2)}+\dots+x^a+1)$$
A: If $n$ is not prime, you can write it as $ab$; with $a,b\in\mathbb N\setminus\{1\}$. Then$$x^n-1=x^{ab}-1=(x^a)^b-1^b=(x^a-1)\left(x^{a(b-1)}+x^{a(b-2)}+\cdots+1\right)$$and therefore$$1+x+\cdots+x^{n-1}=\frac{x^n-1}{x-1}=(x^{a-1}+a^{a-2}+\cdots+1)\left(x^{a(b-1)}+x^{a(b-2)}+\cdots+1\right).$$
A: If $x = 1$, it's obvious.
For $x>1$, let $n$ be composite, say $n = pq$ with $p, q>1$. In that case, we have
$$
(1+x+\cdots+x^{n-1})(x-1) = x^n-1 = x^{pq}-1\\
= (x^p)^q-1\\
= (1+x+x^p+x^{2p}+\cdots + x^{(q-1)p})(x^p-1)
$$
Thus $x^p-1$ divides $x^n-1$, but is not equal to it. In other words, $\frac{x^n-1}{x^p-1}$ is an integer greater than $1$. We get
$$
\frac{x^n-1}{x^p-1} =\frac{(1+x+\cdots+x^{n-1})(x-1)}{(1+x+\cdots +x^{p-1})(x-1)}\\
= \frac{1+x+\cdots+x^{n-1}}{1+x+\cdots +x^{p-1}}
$$
That final fraction is an integer greater than $1$, and the denominator is an integer greater than $1$, which means that the numerator must be composite.
