Why the map $e^{A+B} = e^{A}e^B$ if $A,B$ are matrices that commute

Let $A,B$ be matrices with dimension $N$. Define $ e^A:= I+A+\frac{A^2}{2!}+.... = \lim_{n \to\infty} \sum_{k=0}^n\frac{A^k}{k!}.$

Prove using limits if $AB = BA$ then $ e^{A+B} = e^Ae^B.$ I have observe several answers, but they are too quick and I couldn't understand them.

My first try is to prove $ \sum_{k=0}^{2n}\frac{(A+B)^k}{k!} -\sum_{k=0}^n\frac{A^k}{k!} \sum_{k=0}^n\frac{B^k}{k!} \to 0,$ and then push $n$ to infinity to obtain the promised result.
However, I got stuck in bounding the coefficient of $A^{p}B^{q}$ so I'm not sure if this is a good try.

Edit: In this post, I'm looking for a simple proof using limiting arguments, not ODE proofs or anything like that.