There are familiar functions whose derivatives are periodic. $e^x$ has a period of $1$, since
$$\frac{d}{dx} e^x = e^x$$
And $\sin(x)$ has a period of $4$. I am interested in finding a sequence of functions $f_n$ such that for any $n$, we have
$$ \frac{d^{n}}{dx^{n}}f_n = f_n $$
But for any $0<m<n$
$$ \frac{d^{m}}{dx^{m}} f_n \neq f_n $$
This is just a fancy way of asking if we can find functions that have a period of $n$ with respect to the derivative, for any $n$. We can easily construct such a sequence using a taylor series, by setting
$$f_n= \sum_{k=0}^{\infty}\frac{x^{kn}}{(kn)!} $$
(Note that $f_1(x) = e^x$)
But I am interested in a solution that doesn't rely on the use of infinite series. Does anyone have am interesting solution?
Here's something to notice before we move into the general solution: you mention that $e^x$ has period 1 and $\sin x$ (and $\cos x$) have period 4. Note that \begin{align*} \cos x + i \sin x &= e^{ix} & \cos x - i \sin x &= e^{-ix} \end{align*} Note also that $e^{-x}$ has period 2. We see how for periods 1, 2, and 4, the functions we seek are all of the form $e^{\alpha x}$.
How can we take this observation to find the functions with period $n$? Suppose our function is $f(x) = e^{\alpha x}$. Then we have $$\frac{d^n}{dx^n} f(x) = \frac{d^n}{dx^n} e^{\alpha x} = \alpha^n e^{\alpha x} = \alpha^n f(x)$$ When is it the case that $\frac{d^n}{dx^n} f(x) = f(x)$? Precisely when $\alpha^n = 1$. There are $n$ complex numbers $\alpha$ that satisfy $\alpha^n = 1$; those are numbers of the form $$\alpha = \cos \left(\frac{2k\pi}n \right) + i \sin\left( \frac{2k\pi}n \right)$$ where $k=0, \ldots, n-1$. We have \begin{align*} &e^{\left( \cos \left(\frac{2k\pi}n \right) + i \sin\left( \frac{2k\pi}n \right) \right) x} \\ &= e^{\cos \left(\frac{2k\pi}n \right)x} \cdot e^{i \sin\left( \frac{2k\pi}n \right)x} \\ &= e^{\cos \left(\frac{2k\pi}n \right)x} \left( \cos \left( \sin\left( \frac{2k\pi}n \right)x \right) + i \sin \left( \sin\left( \frac{2k\pi}n \right)x \right)\right) \end{align*} and \begin{align*} &e^{\left( \cos \left(\frac{2(n-k)\pi}n \right) + i \sin\left( \frac{2(n-k)\pi}n \right) \right) x} \\&= e^{\cos \left(\frac{2k\pi}n \right)x} \cdot e^{i \sin\left( \frac{2k\pi}n \right)x} \\&= e^{\cos \left(\frac{2k\pi}n \right)x} \left( \cos \left( \sin\left( \frac{2k\pi}n \right)x \right) - i \sin \left( \sin\left( \frac{2k\pi}n \right)x \right)\right) \end{align*} so in fact the functions (obtained by adding and subtracting the previous two functions) \begin{align*} f_k^+(x) &= e^{\cos \left(\frac{2k\pi}n \right)x} \cos \left( \sin\left( \frac{2k\pi}n \right)x \right) & f_k^-(x) &= e^{\cos \left(\frac{2k\pi}n \right)x} \sin \left( \sin\left( \frac{2k\pi}n \right)x \right) \end{align*} each have period $n$. Here $k$ ranges from $0$ to $\lfloor n/2 \rfloor$ (we don't go all the way to $n$ because $f_{n-k}^+ = f_k^+$ and $f_{n-k}^- = -f_k^-$, so to go beyond $n/2$ would be to repeat functions). Note that $f_0^- = 0$, and if $n$ is even, then $f_{n/2}^- = 0$. As such, the functions $f_k^{\pm}$ comprise $n$ distinct functions. Note that if $f$ and $g$ each have period $n$, then so do $Af + Bg$ where $A$ and $B$ are constants. As such, any linear combination of the aforementioned functions has period $n$ (more precisely, it has period dividing $n$, but it satisfies that it is equal to its own $n$th derivative in any case).
Note that the functions you're looking for are in fact the solutions to the differential equation $$f^{(n)} - f = 0$$ This is an $n$th order linear homogeneous ordinary differential equation, so its solution space has dimension $n$. Indeed, then, we have covered all solutions.