Consider the following two (relatively less common) definitions of the exponential function $e^x: \mathbb{R} \rightarrow \mathbb{R}$:
- Define $e^x$ by the limit:
$$ e^x = \lim_{n \to \infty} \left( 1 + \frac{x}{n}\right)^n $$
- Define $e^x$ as the unique number $y > 0$ such that $$ \int_1^y \frac{dt}{t} = x $$
Question: Do these definitions extend to the complex plane as well (that is, do they make sense if we let $x = z \in \mathbb{C}$)?
As noted by Doug M in the comments above, the binomial expansion of your first example is neatly given as $\sum_{n=0}^\infty\frac{x^n}{n!}$, which is the Taylor expansion of $e^x$ by definition. This expansion analytically extends to any value of $x\in\mathbb C$. Since the limit converges for all $x\in\mathbb C$, it must converge to $e^x$.
Your second example, however, only works for $y>0$, and so by nature it cannot extend to $x\in\mathbb C$. The reason is because the integral of $\frac1t$ is not defined over $t=0$. (as you probably noted) If you did attempt to calculate around the singularity, you would get false statements like $e^0=-1$.