Proceeding directly from the definitions, we have
\begin{equation}
e^0
= \sum_{n = 0}^{\infty} \frac{0^n}{n!}
= 0^0 + \sum_{n = 1}^{\infty} \frac{0^n}{n!}
= 1 + 0
= 1
\end{equation}
since $x^0 = 1$ for every $x \in \mathbb{R}$ and $0^n = 0$ for every integer $n \geq 1$. $\square$
