for a monomial ideal $I$ generated in degree $2$ one has $\index(I^{k+1})>\index(I^k)$ if $\index(I)>1$.

\[maxlinear\] Let $I$ be the edge ideal of a simple graph $G$ and suppose that $I$ has maximal finite index $>1$. Then $\index(I^k)=\infty$ for all $k\geq 2$.

We may assume that $G$ has no isolated vertices. By Theorem \[main2\] we know that $G$ is the complement of an $n$-cycle with $n\geq 5$, in particular $G$ is gap free. We claim that $G$ is claw free. Then by a theorem of Banerjee [@B Theorem 6.17], the assertion follows. In order to prove the claim, let $\{i,i+1\}$, $i=1,\ldots,n-1$ and $\{1,n\}$ be the edges of the cycle $\bar{G}$. Suppose $G$ admits a claw. Then by symmetry we may assume that $\{1,i\}, \{1,j\}$ and $\{1,k\}$ with $1<i<j<k$ are the edges of the claw. However, $\{i,k\}\in \bar{G}$, a contradiction.

Squarefree powers
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Let $I\subset S$ be a squarefree monomial ideal. Then the [*$k$-th squarefree power*]{} of $I$, denoted by $I^{[k]}$, is the monomial ideal generated by all squarefree monomials in $G(I^k)$.

Let $J$ be an arbitrary monomial ideal and let $\alpha=(a_1,a_2,\ldots,a_n)$ be an integer vector with $a_i\geq 0$. Then we let $J_{\leq \alpha}$ be the monomial ideal generated by all monomials 