Over an algebraically closed field of characteristic $0$, every element of finite order acts diagonalisably in a finite-dimensional representation

Let $m=\mathrm{ord}(g)$ and let $H=⟨g⟩$, so $H$ is a finite cyclic group of order $m$. Restrict $\rho$ to $H$. Since $\mathrm{char}k=0$, the integer $m$ is nonzero in $k$, so [L1] makes the restricted representation completely reducible. Also [L3] and [L2] show that $k$ is a splitting field for the finite group $H$, and $H$ is abelian. Therefore [L4] makes every irreducible $H$-summand one-dimensional.

Choose a basis of each one-dimensional $H$-summand and concatenate these bases. Each summand is invariant under $\rho (g)$ and one-dimensional, so $\rho (g)$ acts on it by a scalar. Hence the matrix of $\rho (g)$ in the concatenated basis is diagonal. Therefore $\rho (g)$ is diagonalisable over $k$.