When a function can be expanded as a Taylor series at $x=x_{0}$ we usually write that:
$$f(x)=\sum_{n=0}^{∞}\frac{f^{(n)}(x_0)(x-x_0)^n}{n!}$$
So my question about writing the Taylor series this way is that when $x=x_{0}$ we get that the partial sum of the series for $n=0$ is $f(x_{0})*0^{0}$ , but isn't $0^{0}$ undefined? So my question is shouldn't we clarify that for $n=0$ the partial sum is $f(x_{0})$?
I know my question does not have any mathematical importance but i had this question for a long time about this notation
$0^0=1$ is the usual setting for it makes formulas like the Taylor series more convenient to write. An extra case for $n=0$ is therefore not necessary.
It also makes sense since $0^0$ is an empty product and $1$ is the neutral element of multiplication. Empty sums are equal to zero, empty products are equal to one. In rare cases, $0^0$ is defined as zero, but this is not the norm and should be explicitly stated if such a convention is used. If not, you may assume that the convention is $0^0=1.$