Value of $\sum_{n=0}^{+\infty}\frac{x^n}{n!}$ for $x=0$ Why the value of $\sum_{n=0}^{+\infty}\frac{x^n}{n!}$ for $x=0$, is 1 and not 0?
 A: It's intuitive that the sum of no numbers is defined as zero. Less intuitive is that the product of no numbers is defined as one.
Symbolically:
$$\sum_{\text{false}} x = 0$$
$$\prod_{\text{false}} x = 1$$
No matter what number $x$ is, even zero.
One reason $0^0 = 1$ by definition is because the cardinality (how many elements there are in a set, denoted $\vert \cdot \vert$) of the number of functions from a set $X$ to a set $Y$, denoted $Y^X$, is given by:
$$\left\vert Y^X \right\vert = |Y|^{|X|}$$
and there is one and only one function from the empty set to itself:
$$1 = \left\vert \varnothing^\varnothing \right\vert = |\varnothing|^{|\varnothing|} = 0^0$$
See https://proofwiki.org/wiki/Zero_to_the_Power_of_Zero for more reasons.
Edit: In the context of taking limits, we refer to $0^0$ as an indeterminate form because you cannot conclude $x^y \to 1$ just because $x \to 0$ and $y \to 0$.
A: Because when dealing with power series it is assumed that $0^0=1$.
A: We have by convention $$\left.\sum_{n=0}^{+\infty}\frac{x^n}{n!}\right|_{x=0}=\left.\frac{x^0}{0!}\right|_{x=0}+0+0+0+\dots=1+0+0+\dots=1.$$
