Diagonalizability of elements of finite subgroups of general linear group over an algebraically closed field

Let $A\in G$ be a matrix. After changing bases, $A$ is in JNF. Let us write $A=D+N$ with $D$ diagonal and $N$ the nilpotent part. Now since $G$ is finite, there is some $m\in\Bbb N$ with $A^m=I$, the identity matrix. Since $DN=ND$, we have $$I = A^m = (D+N)^m = \sum_{k=0}^m \binom mk N^k D^{m-k} = D^m + \sum_{k=1}^m \binom mk N^k D^{m-k} =: D^m+\tilde N. $$ Note that $\tilde N$ is a nilpotent, strictly upper triangular matrix. It follows that $\tilde N=0$ and $D^m=I$. Hence, $$ 0 = \tilde N = N\cdot\left(mD^{m-1}+\sum_{k=2}^{l} \binom mk N^{k-1} D^{m-k}\right) $$ and as the second factor is an invertible upper trianglar matrix (because we are in characteristic zero and $m$ is invertible), so we must have $N=0$. In other words, $A=D$ is already diagonal. In other words, $A$ is diagonal up to changing bases, which means that it is diagonalizable.

Let $K$ be a field. As you have already mentioned, a matrix $A \in \operatorname{M}_n(K)$ is diagonalizable (over $K$) if and only if there exists a polynomial $f(t) \in K[t]$ with $f(A) = 0$ such that $f$ decomposes into pairwise different linear factors over $K$.

For $A \in G$ and $n := |G|$ we have that $A^n = I$, so that $A$ satisfies the polynomial $f(t) := t^n - 1 \in \mathbb{k}[t]$. The polynomial $f(t)$ decomposes into linear factors because $\mathbb{k}$ is algebraically closed. It follows from $\operatorname{char}(\mathbb{k}) = 0$ that the polynomial $f$ is seperable (because $f(t) = t^n - 1$ and $f'(t) = n t^{n-1}$ are coprime), so that $f(t)$ decomposes into pairwise different linear factors. Thus $A$ is diagonalizable.

(To see that $f(t)$ is seperable one can also embed $\mathbb{k}$ into $\mathbb{C}$ because $\mathbb{k}$ is an algebraic closure of $\mathbb{Q} \subseteq \mathbb{C}$. As the roots of unity in $\mathbb{C}$ are pairwise different, the same goes for $\mathbb{k}$.)