$SL_n(k)$ is not divisible over any algebraically closed field $k$ and $n>1$. 

This would be indeed a nice exercise in a Linear Algebra course.  

*Proof:* Let $\zeta$ be a primitive $n$th root of unity. I'll show that there is no $X\in SL_n(k)$ such that $X^n=A$ for   
$$A=\scriptstyle\begin{pmatrix}
\zeta & 1  & &   &  \newline 
& \zeta & 1 &  & \newline
& & \ddots & \ddots & \newline  
& & & \zeta & 1 \newline 
& & & & \zeta 
\end{pmatrix}.$$

Let $X^n=A$ and let $\mu_1,...,\mu_n$ be the eigenvalues of $X$. Clearly $\mu_i^n=\zeta$. Each eigenvector of $X$ is also an eigenvector of $A$ and since $A$ has exactly one eigenvector (up to scalar multiples), we conclude $\mu_1=\ldots=\mu_m=:\mu$. Hence $\det(X)=\mu^n=\zeta\neq 1$. qed.