Factorise $x^n + 1$ Is there a way to factorise $x^n + 1$ 
I thought of doing it like this: $$x^n +1 = (x+1)(x^{n-1} - x^{n-2} + \cdots - x + 1)$$ but can't seem to get anywhere using this method.

If $n$ is even then the graph of $y=x^n+1$ is always above the $x$ axis so $x^n+1=0$ has no real roots.
If $n$ is odd then the graph of $y=x^n+1$ crosses the $x$ axis at $x=-1$ and cannot cross the $x$ axis anywhere else because $x^n+1$ is monotonically increasing. So $x^n+1=0$ has a single real root at $x=-1$, and so $x+1$ is a factor of $x^n+1$. In fact, for odd $n$:
$x^n+1=(x+1)(x^{n-1}-x^{n-2}+x^{n-3}-x^{n-4}+\dots-x+1)$
