Is $0^x$ not analytic?

Assume that we define it as follows:

$$f(x)=0\ (x>0)$$

$$f^{(k)}(x)=0\ (x>0)$$

If it were analytic, I could extend its domain by taylor's theorem:

$$f(x)=\sum\_{n=0}^\infty\frac{f^{(n)}(a)}{n!}(x-a)^n$$

If we have $a>0$, then this reduces down to

$$f(x)=0$$

So why can't we define $0^x=0$ like this for all $x$?

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For clarification, I am not asking what $0^{-1}$ or $0^0$ is, I am asking why we can't have analytic continuation of $0^x$ to any $x$ value.