# $n$th derivative of $f(x)$ using limit definition

After playing around with the limit definition of the derivative for higher order derivatives, I noticed the following odd relationship to determine it for an nth order derivative:

Let $F^n=f(x+nh)$ (is there a way to write this properly as an operator on $f(x)$?), then

$$f^{(n)}(x)=\lim_{h\to0}\frac{(F-1)^n}{h^n}=\left(\lim_{h\to0} \frac{F-1}{h}\right)^n$$

Expanding the middle equality gives it in terms of $f(x+nh)$s. Notice that the inside of the bracket on the RHS is equal to $f'(x)$ (sort of). I have actually proven that this is indeed true using repeated use of L'Hopital's law and proof by induction, but I am unsatisfied. The result in the form above seems almost magical, and makes me think there is a very elegant reason why. Can someone explain why? Based on the result, I suspect umbral calculus gives a nice explanation, but I do not know a lot about it. So can someone explain why the above has such an elegant form?

NOTE: I realise that this is similar to this question, but that question is asking about if it's true; I already know it's true, but I'm asking why it's true based on the magical form above. Also, I don't care that this evaluates certain derivatives that shouldn't exist (but does correctly evaluate those that do), so do not worry about that.

Let $D$ be the derivative operator: $(Df)(x):=f'(x)$. The symbol $F$ can be decomposed into an application of the translation operator $T_h$ to $f$. This operator is related to the Taylor series, the formula is stated $$(T_hf)(x) := f(x+h) = (e^{hD}f)(x).$$ If $e^x$ is substituted by its power series, we can obtain the Taylor series without further ado.
A power of $F$ is simply the iterated application of $T_h$. As multiplication of operators is the same as composition of them, we have $F^n=(T_h^nf)(x)$.
Now, by analogy $$D = \ln(e^D) = \lim_{h\rightarrow 0} \frac{(e^D)^h-1}{h}$$ and therefore $$D^n = \left[\lim_{h\rightarrow 0} \frac{e^{hD}-1}{h}\right]^n = \lim_{h\rightarrow 0} \left[\frac{e^{hD}-1}{h}\right]^n.$$ To be precise, the one in $e^{hD}-1$ is the identity operator $I$ which maps a function to itself, i.e. $(If)(x)=f(x)$.