Maschke's Theorem

Step 1 (start with any complement). By ordinary linear algebra, choose any vector-space complement $U$ to $W$ in $V$ (not necessarily $G$-invariant), and let $\pi: V \to W$ be the corresponding projection ($\pi|_W = \mathrm{id}$, $\ker\pi = U$).

Step 2 (average over the group). Define $\bar\pi = \dfrac{1}{|G|}\sum_{g \in G} \rho(g)\, \pi\, \rho(g)^{-1}$. This makes sense exactly because $\operatorname{char}(k) \nmid |G|$, so $|G|$ is invertible in $k$.

Step 3 ($\bar\pi$ is still a projection onto $W$). For $w \in W$, since $W$ is $G$-invariant, $\rho(g)^{-1}w \in W$, so $\pi(\rho(g)^{-1}w) = \rho(g)^{-1}w$, giving $\rho(g)\pi(\rho(g)^{-1}w) = w$ for every term; averaging these $|G|$ copies of $w$ gives $\bar\pi(w) = w$. For general $v \in V$, $\pi(\rho(g)^{-1}v) \in W$ always (image of $\pi$), so $\rho(g)\pi(\rho(g)^{-1}v) \in \rho(g)W = W$ by invariance, and averaging stays in $W$; hence $\bar\pi: V \to W$ is a projection onto $W$.

Step 4 ($\bar\pi$ commutes with the action, so its kernel is $G$-invariant). For any $h \in G$, $\rho(h)\bar\pi\rho(h)^{-1} = \dfrac{1}{|G|}\sum_{g} \rho(hg)\pi\rho(hg)^{-1} = \bar\pi$ since $g \mapsto hg$ merely permutes the sum; so $\rho(h)\bar\pi = \bar\pi\rho(h)$. Setting $W' = \ker\bar\pi$, for $v \in W'$ we get $\bar\pi(\rho(h)v) = \rho(h)\bar\pi(v) = \rho(h)(0) = 0$, so $\rho(h)v \in W'$: $W'$ is $G$-invariant, and $V = W \oplus W'$ since $\bar\pi$ is a projection onto $W$. Iterating this splitting on $W'$ (finite dimension forces termination) decomposes $V$ into a direct sum of irreducibles.