Special linear group over algebraically closed field is divisible precisely by those primes that do not divide its degree

Statement
Suppose $$K$$ is an algebraically closed field and $$n$$ is a natural number. Consider the special linear group $$SL(n,K)$$. Suppose $$p$$ is a prime number. Then, $$SL(n,K)$$ is $$p$$-divisible if and only if $$p$$ does not divide $$n$$.

Related facts

 * General linear group over algebraically closed field is divisible
 * Derived subgroup not is divisibility-closed

Not divisible if $$p$$ divides $$n$$
Suppose $$p^r$$ is the largest power of $$p$$ dividing $$n$$.

Let $$\zeta$$ be a primitive $$(p^r)^{th}$$ root of unity, and consider the Jordan block of size $$p$$ for the eigenvalue $$\zeta$$:

$$g = \begin{pmatrix} \zeta & 1 & 0 & 0 & \dots & 0 \\ 0 & \zeta & 1 & 0 & \dots & 0 \\ \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\\cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\ 0 & \dots & 0 & 0 & \zeta & 1 \\ 0 & \dots & 0 & 0 & 0 & \zeta \\\end{pmatrix}$$

Suppose $$x^p = g$$. Then, $$x$$ must commute with $$g$$, and hence, all its eigenvalues must be $$p^{th}$$ roots of eigenvalues of $$g$$. Note also that $$x$$ must be a single Jordan block (because if it were multiple Jordan blocks, then all its powers would also be multiple Jordan blocks). This forces that $$x$$ should have a single eigenvalue with multiplicity $$n$$, and that this must be a $$(p^{r + 1})^{th}$$ root of unity. The determinant of any such element would then be a primitive $$p^{th}$$ root of unity and would in particular not be $$1$$, so $$x$$ would not be in $$SL(n,K)$$.

Divisible if $$p$$ does not divide $$n$$
There must be at least one Jordan block whose size is not a multiple of $$p$$. Take $$p^{th}$$ roots willy-nilly of the other Jordan blocks. For this one Jordan block, choose an appropriate $$p^{th}$$ root to get a determinant of 1. The existence of such a $$p^{th}$$ root can be worked out from the Chinese remainder theorem.