Let $M$ be a finitely generated $F$-module where $F$ is a field. Let $\{m_1,....m_n\}$ be the generating set for $M$.
Then, given $x \in M$, $\exists r_1,r_2,...r_n \in F$ such that $r_1m_1+....+r_nm_n=x$.
If $M$ is a completely reducible module, then given any submodule $N \leq M$, $\exists Z \leq M$ such that $M = Z \oplus N$.
So, back to the original statement. If $M$ is a module over a field, then is $M$ a direct product of copies of $F$? Why is this? If this is so, then the statement is simple, since any submodule of $M$ is just a direct product of some of the summands that compose $M$, and the summands that you don't use form a submodule of $M$ that is the complement of this first one.
$F$ is a simple $F$ module, and by basic linear algebra, every $F$ module is isomorphic to one of the form $\oplus_{i\in I}F$ for some index set $I$.