I was trying to find $0!$ using the series expansion of $e^x$ and got some odd results.
We know that, $$e^x = \sum_{n =0}^\infty\dfrac{x^n}{n!}$$
Putting $x = 0$ will give us following results.
$$e^0 = \sum_{n=0}^\infty \dfrac{0^n}{n!}\implies1 = \dfrac{0^0}{0!} +\underbrace{ \dfrac{0^1}{1!} + \dfrac{0^2}{2!} + ...\infty}_0$$
So, we are ended up with $1 = \dfrac{0^0}{0!}$
On cross multiplication, we have $0! = 0^0$ (I'm not sure if cross multiplication is a valid step here).
What to do with this? Can we conclude that $0^0 = 1$ because $\lim\limits_{x\to0^+} x^x = 1?$
Is the step of cross multiplication really a valid step above?
I would say: facts $(1)$ and $(2)$ are are used in the proof of fact $(3)$. Therefore, it would be "circlular" to deduce $(1)$ or $(2)$ from $(3)$.
$$ x^0 = 1\quad\text{where $x$ is a real variable} \tag1$$
$$ 0! = 1 \tag2$$
$$ e^x = \sum_{n=0}^\infty \frac{x^n}{n!} \tag3$$
I stated $(1)$ in this peculiar way because that formulation is more widely accepted than $$ 0^0=1 \tag{1'}$$