Not sure what they want here, please help.
show funcion f(x) = n=0->infinity x^n / n!
show that f(x) = e^x
Depends what you can use. If you can just use a Taylor series this is straightforward...
A Taylor's series is
$\displaystyle f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}(x-a)^n.$
where $\displaystyle f^{(n)}(a)$ means the nth derivative of f(x) evaluated at x=a. So in your case
$\displaystyle f^{(n)}(x) = e^x$
and you substitute a=0 to get the result you need.
Note: A Taylor series with a=0 is also called a Maclaurin series.
Hope this helps.