What's next in this sequence of polynomials?

x1

x+1

x^{2}+x+1

x^{2}+1

x^{4}+x^{3}+x^{2}+x+1
These are the first 5 cyclotomic polynomials. The nth
cyclotomic polynomial is
cyc_{n}(x)=(xe_{1})(xe_{2})...(xe_{r})
where e_{1},e_{2},...,e_{r}
are the primitive complex
nth roots of unity.
It follows from the definition that the degree of cyc_{n}
is phi(n), where phi is the Euler Phi function.
In the case of a prime number p we get
cyc_{p}(x)=(x^{p}1)/(x1)=x^{p1}+x^{p2}+...+x+1
(because all the pth roots of unity are primitive except for 1).
The cyclotomic polynomials can be calculated recursively
because of the formula
x^{n}1=product over all dn of cyc_{d}(x)
This formula follows because the roots
of x^{n}1 are exactly the nth roots of
unity. By Lagrange's Theorem each nth root of unity
is a primitive dth root of unity for some divisor d of n. On the other hand clearly any such primitive dth root of unity is an nth root of unity.
Let's do this for an example, n=6. The divisors of 6 are
1,2,3,6 so the formula tells us
x^{6}1=cyc_{1}(x)cyc_{2}(x)cyc_{3}(x)cyc_{6}(x)
=(x1)(x+1)(x^{2}+x+1)cyc_{6}(x)
=(x^{2}1)(x^{2}+x+1)cyc_{6}(x)
=(x^{4}+x^{3}x+1)cyc_{6}(x)
We obtain from this that cyc_{6}(x)=x^{2}x+1.
A couple more interesting properties of cyc_{n}(x)