= Solution
In a submultiplicative <matrix norm>, the exponential series is absolutely convergent. Its <Cauchy product> can therefore be grouped by the total degree:
$$
e^Ae^B=\sum_{n=0}^{\infty}\sum_{j=0}^n\frac{A^jB^{n-j}}{j!(n-j)!}
=\sum_{n=0}^{\infty}\frac1{n!}\sum_{j=0}^n\binom nj A^jB^{n-j}.
$$
When $AB=BA$, the inner sum is $(A+B)^n$ by the <binomial theorem>. Thus $\boxed{e^Ae^B=e^{A+B}}$. The commutation assumption is used precisely in that binomial expansion; absolute convergence alone does not supply it.
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