When looking at this problem, I found the series to be:
$$\sum_{n=0}^{\infty}\frac{{x^{40n}}}{n!}$$
When finding the $20$th and the $80$th derivative using this series, shouldn't they all be zero? I confirmed that the $20$th derivative is but the $80$th derivative isn't.
Is there a reason why?
As @Clement C mentions, when you differentiate $80$ times the term at $0$ doesn't cancel. Another perspective is taking the Maclaurin series of $e^x$, i.e. $$e^y = \sum_{n=0}^{\infty} \frac{y^n}{n!}$$ Now let $y = x^{40}$, $$e^{x^{40}} = \sum_{n=0}^{\infty} \frac{{x^{40n}}}{n!}$$ which tells you that the exponents are multiples of $40$, i.e. everything in between is zero.