By application of the exterior derivative, a $p$-form $w$ becomes a $p+1$ form. A form must be exact if it is the exterior derivative of another form.
Consider a $0$-form $w$ on $\mathbb{R}^n$, this is just a function of $n$ variables. By the first two points above, this $0$-form for arbitrary $n$ must be non-exact since $p$ forms where $p<0 $ does not exist?
I am new to differential forms so any feedback is much appreciated
There is exactly one exact $0$-form, by convention, and that's the function which evaluates to $0$ everywhere. This convention is entirely a convenience. As you say, if one is to be consistent in this matter, then no $0$-form is exact. However, the resulting theory becomes nicer if we let the zero function be exact, so that's what we do.