Is $\mathrm{GL}_n(K)$ divisible for an algebraically closed field $K?$ This is a follow-up question to this one. To reiterate the definition, a group $G$ (possibly non-abelian) is divisible when for all $k\in \Bbb N$ and $g\in G$ there exists $h\in G$ such that $g=h^k.$ Let $K$ be an algebraically closed field. For which $n$ is $\mathrm{GL}_n(K)$ divisible? (It is clearly true for $n=0$ and $n=1$.) The linked question is about the case of $K=\Bbb C$ and the answer is "for all $n$" there.

${\rm GL}(n,K)$ is not divisible when $K$ has finite characteristic $p$ and $n >1.$ The maximum order of an element of $p$-power order in ${\rm GL}(n,K)$ is $p^{e+1},$ where $p^{e} < n \leq p^{e+1}$ ($e$ a positive integer). There is an element of that order, and it is not the $p$-th power of any element of ${\rm GL}(n,K).$
