Really interesting read, thank you! One thing I found especially interesting: They argue that you can express the (complex) exponential function exp(x) as a po

Really interesting read, thank you! One thing I found especially interesting: They argue that you can express the (complex) exponential function exp(x) as a power series with powers x^k. They do not say it explicitly, but if we assume the power series comes from a Taylor series, then the k-th factor 1/(k!) is the derivative evaluated at x=0. And the k-th derivative has exactly the unit needed to cancel the unit of x^k. So, all summands of the series are unitless and hence the exponential function's argument is unitless. This argument would hold for any function which can be written as a series like this. I am wondering whether this is actually a "problem" of the derivative operator.