Let $\mathbb{K}$ be an algebraically closed field and $V$ an n-dimensional $\mathbb{K}-$vector space. Suppose $G \leq GL(V)$ is completely reducible and that for some $d \in \mathbb{N}$ we have $ g^d = id, \forall g \in G$. Prove that $\vert G \vert \leq d^{n^3}$.
What I've done so far:
First, assume that $G$ is irreducible. A version of Burnside's Theorem tells us that the set $G$ spans $V$ as a $\mathbb{K}-$vector space, so $G$ contains $m = n^2$ linearly independent elements, say $w(1),...,w(m)$. Take an $m-$tuple $\mu \in \mathbb{K}^m$ and define $$G(\mu) = \{g \in G \mid trace(gw(k)) = \mu_k, \forall k \in \{1,...,m\}\}.$$ This gives $m$ linearly independent equations for $m$ variables, so there is at most $1$ solution $(g_{ij})$. Observe that if $\lambda$ is an eigenvalue of $g \in G$, then the condition $g^d = 1$ implies $\lambda^d = 1$, for which there are $d$ different possibilities ($\mathbb{K}$ is algebraically closed). Since the trace is equal to the sum of the $n$ eigenvalues, we have that there are only $(d^n)^{m}=d^{n^3}$ possibilities for $\mu$, and so $\vert G \vert \leq d^{n^3},$ as required. Now, assume that $V \cong \bigoplus_{i \in I}V_i$ for some (finite) family $I$ with $\vert I \vert \geq 2$, where each $V_i$ is a subspace of V. If we let $G_i = G \vert_{V_i}$ be the restriction of $G$ to $V_i$ and $n_i = dim(V_i)$, then the same argument tells us that $\vert G_i \vert \leq d^{nn_i^2}$. Finally, we must have $\vert G \vert \leq \prod_{i \in I}\vert G_i \vert = d^{n \sum n_i^2} < d^{n^3},$ using that $\sum n_i^2 < \left(\sum n_i\right)^2 = n^2$.
I'm pretty sure the part where $G$ is irreducible is correct, but what about the rest?
Thanks in advance!