Combination of linear functions that give the derivative operator Let $D$ be the derivative operator and $C^\infty$ the set of functions
  derivable infinitely many times.
  
  Here $f^n=f\circ f\circ\cdots\circ f\text{, }n\text{ times}$


It can be easily shown that there exists no linear function $f\in\mathcal{L}(C^\infty)$  such that $f^2=D$.

If I am not wrong, this can be easily extended to the non existence of a function $f\in\mathcal{L}(C^\infty)$ such that $\prod f=D$ as a finite product.

What can be said for an infinite product? How can one (dis)approve of the existence of a linear function $f\in\mathcal{L}(C^\infty)$ such that $f^n=D$ ?

It seems unlikely to me, but I have found no way to prove it.
Not claiming to understand the main question of the "infinite product", I could comment on the title and the linear non-existence claim. Fractional derivative is a linear operator, where one could write $D^{\frac{1}{2}}$ as a functional square root of the differentiation operator that when applied twice to a function will have the same effect as differentiation. (see further, e.g., the "Short Introduction to Fractional Calculus").
Not claiming to understand the main question of the "infinite product", I could comment on the title and the linear non-existence claim. Fractional derivative is a linear operator, where one could write $D^{\frac{1}{2}}$ as a functional square root of the differentiation operator that when applied twice to a function will have the same effect as differentiation. (see further, e.g., the "Short Introduction to Fractional Calculus").