Looking for a function that fits a certain criteria I am looking for a function that fits this description:
$$
\frac{d^n}{dx^n}[f(x)] = n! f(x)
$$
or
$$
\frac{d^n}{dx^n}[f(x)] = (n-1)! f(x)
$$
For all values of $n$, with this function i am looking to derive a new identity, would be greatly appreciated.
 A: Try solving the system of ODEs and you'll see the only function is trivial.
$$\frac{\mathrm{d}}{\mathrm{d}x} [f(x)] = f(x) \;\; \Rightarrow \;\; f(x) = Ce^t, \,C\in\mathbb{R}$$
$$\frac{\mathrm{d}^2}{\mathrm{d}x^2}[ f(x)] =  \frac{\mathrm{d}^2}{\mathrm{d}x^2} [Ce^t] = Ce^t = 2Ce^t \;\; \Rightarrow \;\; C = 0,$$
i.e. $f(x) \equiv 0$. Clearly $f(x) \equiv 0$ solves the system of ODEs in the first identity so it is a solution. Similarly, we can show that the zero function is the only function satisfying the second system.
A: If $\dfrac{d^n}{dx^n}[f(x)] = n!f(x)$ for all $x \in \mathbb{R}$, then we have: 
$2!f(x) = \dfrac{d^2}{dx^2}[f(x)] = \dfrac{d}{dx}\left[\dfrac{d}{dx}[f(x)]\right] = \dfrac{d}{dx}[1!f(x)] = 1!\dfrac{d}{dx}[f(x)] = (1!)^2f(x)$.
So, $2f(x) = f(x)$, i.e. $f(x) = 0$ for all $x \in \mathbb{R}$. 
Similarly, if $\dfrac{d^n}{dx^n}[f(x)] = (n-1)!f(x)$ for all $x \in \mathbb{R}$, then we have: 
$2!f(x) = \dfrac{d^3}{dx^3}[f(x)] = \dfrac{d^2}{dx^2}\left[\dfrac{d}{dx}[f(x)]\right] = \dfrac{d^2}{dx^2}[0!f(x)] = 0!\dfrac{d^2}{dx^2}[f(x)] = 0! \cdot 1!f(x)$.
So, $2f(x) = f(x)$, i.e. $f(x) = 0$ for all $x \in \mathbb{R}$. 
Hence, the only function that satisfies either identity is $f(x) \equiv 0$, which isn't very interesting.
