Why does $\sum_{i=0}^{\infty}\frac{a^i}{i!}=e^a$? Here is a standard identity:

$$\sum_{i=0}^{\infty}\frac{a^i}{i!}=e^a$$

Why does it hold true?
A brief answer: Let's consider the exponential function $e^x$. The definition of $e$ is that $\frac{d}{dx}e^x = e^x$. Now let's assume that $e^x$ can be written as an infinite sum of the form $\sum_{i=0}^{\infty}a_ix^i$. Using the sum rule for derivatives, we have $\sum_{i=0}^{\infty}a_ix^i = \sum_{i=0}^{\infty}\frac{d}{dx}a_ix^i = \sum_{i=1}^{\infty}ia_{i}x^{i-1}$. Therefore, $a_i = \frac{a_{i-1}}{i}$, so starting with the base case of $i=0$, $a_i = \frac{1}{i!}$.



(Warning: variables change from the previous section) 

Another way to look at this is to consider the more standard definition of $e$, which is $\lim_{n\to\infty}(1 + \frac{1}{n})^n$. Therefore, $e^a$ can be written as $\lim_{n\to\infty}(1 + \frac{1}{n})^{an}$. Using the binomial theorem, the expression expands to $$\lim_{n\to\infty}(1 + \frac{an}{n} + \frac{an(an - 1)}{2n^2}+ \frac{an(an-1)(an-2)}{6n^3} + \ldots)$$ The lesser powers in the numerators drop out, so the expression becomes $$1 + a + \frac{a^2}{2} + \frac{a^3}{6} + \ldots$$which is $\sum_{i=0}^{\infty}\frac{a^i}{i!}$ in sum notation.
The Taylor series for $e^x$ is

$$
\sum_{k = 0}^{\infty} \frac{x^k}{k!}.
$$

Letting $x = a$, we obtain

$$
e^a = \sum_{k = 0}^{\infty} \frac{a^k}{k!}
$$

as desired.