Why $x^4+x^3+x^2+x+1$ has root order 1 or 5 in $GF(2)$?
For example: $P(x)=x^4+x^3+x^2+x+1$ is irreducible over $GF(2)$. Note, that $(x^5-1)=(x-1)(x^4+x^3+x^2+x+1)$. Those, any root of $P(x)=0$ has order 5 or 1.
Here the order of a root $\alpha$ clearly means the order of $\alpha$ in multiplicative group $GF(2)[\alpha]^*$. IOW, the order of $\alpha$ is the smallest positive integer $m$ such that $\alpha^m=1$. So let us assume that $P(\alpha)=0$, where $\alpha$ belongs to some extension field $GF(q)$, $q$ a power of two.
The factorization that you gave $$ x^5-1=(x-1)P(x) $$ shows that $$ \alpha^5-1=(\alpha-1)P(\alpha)=(\alpha-1)\cdot0=0. $$ Therefore $\alpha^5=1$. From the basic theory of cyclic groups we know that the order of $\alpha$ must then be a factor of five, i.e. either five or one. The latter possibility is easy to exclude. If $\alpha$ were of order one, then $1=\alpha^1=\alpha$. But $P(1)=5=1\neq0$, so $1$ is not a zero of $P(x)$.