$SL_n(k)$ 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.

$SL_n(k)$ 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 $\zeta\neq 1$. 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 $\mu_1=\ldots=\mu_n$. Hence $\det(X)=\mu_1^n=\zeta\neq 1$. qed.