Domain of a Polynomial function

This is because $0^0$ is indeterminate.

This is an extremely common misconception. There is a vast difference between $0^0$ and the form of a limit, which may be labelled as "$0^0$" (note the quotes!), just as there is a difference between $\frac00$ and the form "$\frac00$" of some limits.

Here are the facts under standard mathematical conventions:

$0^0 = 1$ in contexts where the exponent is a natural number.

"$0^0$" is a label referring to an indeterminate form of some limits.

$\frac00$ is undefined.

"$\frac00$" is a label referring to another indeterminate form of some limits.

Limits with form "$0^0$" or "$\frac00$" may have a value or may not. That is precisely why we call their form indeterminate, because we cannot determine the value so easily by their form alone.

$0^0$ is not a limit, and if the exponent is a natural number (like for rings or in combinatorics or in the binomial theorem or in power series or ...) then its value is always $1$.

If you do not believe this, see the conventional statement of the binomial theorem here and here (equation 4) and the definition of power series here and here.

On pondering this good question further, I think that part of the problem is that we have no name for the functions $x\mapsto x^n$. A clean way of getting around the difficulty might be the following:

Define functions $P_n$ for nonnegative integers $n$ inductively as follows: for all $x$, $P_0(x)=1$, and for $n\ge0$, define $P_{n+1}(x)=xP_n(x)$. You see that this makes $P_0$ the constant function $1$, and for $n>0$, $P_n(x)=x^n$.

Then your function can be written $\sum_{i=0}^na_iP_i\>$.

Formally, you are absolutely correct. $0^0$ is an indeterminate form. But consider a seemingly unrelated case:

$$f(x)=\frac{x}{x}.$$

This used to drive me nuts, because it is clearly just the same as the function $g(x)=1$... right? The answer is no, but only in a way that is disgustingly technical. Similar to your case, $f(0)$ is technically an indeterminate form. The problem is division of $0$ by $0$. So the functions $f(x)$ and $g(x)$ can't really be equal because they have different domains. However, there is a way around this. Consider instead defining a new function $h$ in this way:

$h(x)= \frac{x}{x}$ if $x\neq 0$, and $h(x)=1$ if $x=0$. Now, we have removed the problem with $0$ and defined a function truly equal to $g(x)=1$ everywhere.

In your problem, $a_0$ is not really equal to $a_0x^0$ because those expressions have different domains. Specifically, $0$ is in the domain of the first, but not of the second. However, when people speak of $a_0$ as being "the $0$-order term", they are doing that for reasons that are intuitively helpful, not reasons that are mathematically formal. And it is always helpful to remember that $a_0\neq a_0x^0$ in general, but that $a_0= a_0x^0$ when $x\neq 0$.

Does this help? Please do ask for clarification if you need it, as this is not only an important point but demonstrates great mathematical insight on your part. I enjoyed thinking about it.