Fundamental question about dimensional analysis

A standard argument to deny possibility of inserting dimensionful quantities into transcendental functions is the following expression for Taylor expansion of e.g. $\exp(\cdot)$:

$$ e^x = \sum_n \frac{x^{n}}{n!} = 1 + x +\frac{x^2}{2} + \dots\,.\tag1$$

Here we'd add quantities with different dimensions, which you have already accepted makes no sense.

OTOH, there's an argument (paywalled paper), that in the Taylor expansion where the derivatives are taken "correctly", you'd get something like the following for a function $f$:

\begin{multline} f(x+\delta x)=f(x)+\delta x\frac{df(x)}{dx}+\frac{\delta x^2}2\frac{d^2f(x)}{dx^2}+\frac{\delta x^3}{3!}\frac{d^3f(x)}{dx^3}+\dots=\\ =f(x)+\sum_{n=1}^\infty\frac{\delta x^n}{n!}\frac{d^nf(x)}{dx^n},\tag2 \end{multline}

and the dimensions of derivatives are those of $1/dx^n$, which cancel those of $\delta x^n$ terms, making the argument above specious.

(I know I am answering an old question, but I think the following is a nice way to explain to young students.)

You don't need to know Taylor expansions. Simply remember the definition of the exponential. It satisfies the differential equation

$$ \frac{\text d y}{\text d x} = y(x) $$

According to this, the derivative of $\text e^x$ has the same dimension as $\text e^x$. Therefore, $x$ should be dimensionless, since the derivative of $\text e^x$ has the dimension of $\text e^x$ divided by $x$. (This assertion comes from the definition of the derivative as a limit and it is also suggested by the $\text d / \text d x$ notation.)

Because of the way an exponential is defined. By an expression like $a^b$ we mean to say that the quantity $a$ is multiplied $b$ times with itself. Thus an expression like $(5m)^{7s}$ would mean $5m$ multiplied "7 seconds" times with itself, which is meaningless.