  Every finite graph admits a \emph{simple (topological) drawing}, that is, a
drawing where every pair of edges intersects in at most one point. However, in
combination with other restrictions simple drawings do not universally exist.
For instance, \emph{$k$-planar graphs} are those graphs that can be drawn so
that every edge has at most $k$ crossings (i.e., they admit a \emph{$k$-plane
drawing}). It is known that for $k\le 3$, every $k$-planar graph admits a
$k$-plane simple drawing. But for $k\ge 4$, there exist $k$-planar graphs that
do not admit a $k$-plane simple drawing. Answering a question by Schaefer, we
show that there exists a function $f : \mathbb{N}\rightarrow\mathbb{N}$ such
that every $k$-planar graph admits an $f(k)$-plane simple drawing, for all
$k\in\mathbb{N}$. Note that the function $f$ depends on $k$ only and is
independent of the size of the graph. Furthermore, we develop an algorithm to
show that every $4$-planar graph admits an $8$-plane simple drawing.
