Exponential series representation It is well known that the exponential function can be represented as follows:
$$e^x=\sum_{n=0}^\infty\frac{x^n}{n!}$$
However, the right hand side is not defined at $x=0$ due to $n=0$. Is it wrong? should we use  this one:
$$e^x=1+\sum_{n=1}^\infty\frac{x^n}{n!}\text{?}$$
 A: First of all, $0!=1.$ This is not in dispute. It is just the definition. There are reasons for the definition, but I won’t go into them here.
This example is one of many reasons to define $0^0=1.$ There are several other reasons, but there are other questions covering that on this site.
I’ll just add that there is no problem with defining $0^0=1.$ It doesn’t lead to paradoxes, just discontinuities.
That said, even if we left $0^0$ undefined, we’d still treat $e^0=1$ using the standard power series notation. The notation is too convenient.
Do we really want to write:

If $$f(x)=a_0+\sum_{n=1}^\infty a_nx^n$$ then $$f’(x)=1\cdot a_1+\sum_{n=1}^\infty (n+1)a_{n+1}x^n?$$

That hides the pattern. You have to read carefully to realize the constant term follows the pattern of the other terms.
It would be notation that obscures rather than clarifies.
A: In this context one should define $0^0=1$ and $0!=1$ because $\displaystyle \prod_{x\,\in\,\varnothing} x = 1.$
Proof:
$$
\prod_{x\,\in\,A} x = \prod_{x\,\in\,A\,\cup\,\varnothing} x = \prod_{x\,\in\,A }x \cdot \prod_{x\,\in\,\varnothing} x.
$$
Now divide both sides by the leftmost expression. $\qquad\blacksquare$
