Let $F:\mathcal{B}\to\mathbb{C^{n\times n}}$, where $\mathcal{B}=\{z\in\mathbb{C}:|z|<R\}$ be given by $F(z)=\sum_{k\geq 0} A_k z^k$, with some radius of convergence $R$. (edit: $A$ is obviously some matrix on $\mathbb{C^{n\times n}}$.)
I have figured out that this function is holomorphic (we have $\frac{F^{(t)}}{t!}=A_t$, the result is almost immediate). Now consider the "differential equation" $F'(z)=AF(z)$.
We can deduce that $A_{k+1}=\frac{AA_k}{k+1}$, and then get the general expression for $A_k=\frac{A^kA_0}{k!}$.
Then, by the definition of $F$, we get $F(z)=\sum_{k\geq 0}\frac{A^kA_0}{k!}z^k$, which sums to $e^{zA}A_0$. My question is: my professor got the answer $A_0e^{zA}$, because on the sum above, he commuted $A_0$ with $A^k$. I'm not really understanding why we can do that. I do know we can commute the terms in $e^AA$, but I think $A^k\neq A_0$? I'd love for a bit of insight into this.
You are correct and the professor is wrong. The solution for the matrix differential equation $F' = A F$ with $F(0) = A_0$ is $F(z) = e^{zA} A_0$, and this is not the same as $A_0 e^{zA}$ unless $A$ and $A_0$ happen to commute.