Commuting in matrix exponential

Question

Let $A$, and $B$ be commuting $n\times n$ matrices, i.e., $A B = B A$. Let $$ \exp(A) := \sum_{i=0}^\infty\frac{1}{i!} A^i $$ Show that $\exp(A+B) = \exp(A) \exp(B)$.

Answer 1

By Cauchy product we have $$\exp(A)\exp(B)=\left(\sum_{i=0}^\infty\frac{1}{i!} A^i\right)\left(\sum_{i=0}^\infty\frac{1}{i!} B^i\right)=\sum_{i=0}^\infty c_i$$ where $$c_i=\sum_{k=0}^i\frac{1}{k!} A^k\frac{1}{(i-k)!} B^{i-k}=\frac{1}{i!}\sum_{k=0}^i{i\choose k}A^k B^{i-k}=\frac{1}{i!}(A+B)^i$$ hence $$\exp(A)\exp(B)=\sum_{i=0}^\infty\frac{1}{i!}(A+B)^i=\exp(A+B)$$