The Taylor series for $e^x = \sum_{i=0}^\infty \frac{x^i}{i!}$. Then as $e^0 = 1$, if one evaluates the Taylor series at $x=0$ we find that $e^0 = \sum_{i=0}^\infty \frac{0^i}{i!} = \frac{0^0}{0!} + \frac{0^1}{1!} + \frac{0^2}{2!} + \ldots = 1$. So we are taking $0^0 = 1$?
Given we define $x^0 = 1, \ \forall x \in \mathbb{R}$ and $0^y = 0, \ \forall y>0 $ what is the justification for giving precedence to the former rule when evaluating the sum?
a) As you noticed $\mathop {\lim }\limits_{\left( {x,y} \right)\; \to \;\left( {0,0} \right)} x^{\,y} $ is indefinite, as it depends on the path followed to approach the origin
b) For fixed $y$, as in Taylor expansion, a "reason" for defining $x^{\,0} = 1\quad \left| {\;\forall x} \right.$ is (to my opinion) that we want to preserve the binomial theorem $$ 1 = \mathop {\lim }\limits_{x\; \to \;0} \left( {1 + x} \right)^{\,y} = \sum\limits_{0\, \le \,k} {\left( \matrix{ y \cr k \cr} \right)x^{\,k} } $$