Why is $\sum^\infty_{n=0}\sum^n_{k=0}\frac{b^k}{k!}\cdot\frac{a^{n-k}}{(n-k)!}=\sum^\infty_{n=0}\frac{(a+b)^n}{n!}$?

The Binomial theorem states that $$\sum_{k=0}^n{n\choose k}b^k a^{n-k}=(a+b)^n,$$ so $$\sum\limits^\infty_{n=0} \sum\limits^n_{k=0} \frac {b^k} {k!} \cdot \frac {a^{n-k}} {(n-k)!}=\sum\limits^\infty_{n=0} \frac{1}{n!}\left(\sum\limits^n_{k=0} \frac{n!}{k!(n-k)!}{b^k}a^{n-k}\right)\\\quad\quad\quad\quad=\sum\limits^\infty_{n=0} \frac{1}{n!}\sum\limits^n_{k=0} {n\choose k}b^ka^{n-k}\\\quad= \sum\limits^\infty_{n=0} \frac {{(a+b)}^n} {n!}.$$