The value of the Nth derivative equals the function. Is there another function where $f\left(x\right)=f^{\left(n\right)}\left(x\right)$ other than $e^x$ ?

Sorry that I wasn't clear before, I was meaning for all possible derivatives. 
 A: First observe, if $f(x) = f^{(n)}(x)$ for all $n \geq 1$ and for all $x$, then $f(x) = f'(x)$.  The answer is actually different if the domain is disconnected.  We only consider $f$ defined on an interval (which is connected of course).  Claim: $f = f'$ if and only if there is a constant $k \in \mathbb{R}$ such that $f(x) = ke^{x}$.  The only non-trivial direction is the "only-if".  Assume $f = f'$ on an interval.
\begin{eqnarray*}
   f' &=& f \\
  f' - f &=& 0\\
  e^{-x}f' - e^{-x}f &=& 0 \\
  (e^{-x}f)' &=& 0
\end{eqnarray*}
Thus, since the interval is connected, we know that $(e^{-x}f) = k$ for some constant of integration $k$.  This yields $f = ke^x$.
A: It depends. If you want $f = f^{(n)}$ for all values of $n$, then no: $f(x) = \operatorname{e}^x$ is the only option. Well, you can have a constant multiple: $\alpha \operatorname{e}^{x}$. 
If you want $f = f^{(n)}$ for a specific, fixed value of $n$, then yes: there are other options. For example, if $n=4$ then $f(x) = \sin x$ and $f(x) = \cos x$ are two possibilities. Let's look at why:
We usually take $f(x) = \operatorname{e}^{\lambda x}$ as a trial function. Then $f'(x) = \lambda\operatorname{e}^{\lambda x}$, $f''(x) = \lambda^2\operatorname{e}^{\lambda x}$, ... , $f^{(n)}(x) = \lambda^n\operatorname{e}^{\lambda x}$. If you want $f=f^{(n)}$ then you want $f^{(n)}(x) - f(x)=0$ for all $x$, i.e.
\begin{array}{ccc}
\lambda^n\operatorname{e}^{\lambda x}-\operatorname{e}^{\lambda x} &\equiv& 0 \\
(\lambda^n-1)\operatorname{e}^{\lambda x} &\equiv& 0
\end{array}
Hence you want $\lambda^n-1=0$. There will be $n$ solutions to this: $\lambda_1,\ldots,\lambda_n$, and the solutions to $f^{(n)}=f$ is given by $\alpha_1\operatorname{e}^{\lambda_1 x} + \cdots + \alpha_n\operatorname{e}^{\lambda_n x}$, where the $\alpha_i$ are constants. Consider the case $n=4$. We need to solve $\lambda^4-1=0$, hence $\lambda=\pm 1, \pm i$ so the solutions are of the form:
$$f(x) \equiv \alpha_1\operatorname{e}^{-x} + \alpha_2\operatorname{e}^{x} + \alpha_3\operatorname{e}^{-ix} + \alpha_4\operatorname{e}^{ix}$$
If I choose $\alpha_1=\alpha_2=0$ and $\alpha_3=\alpha_4=1$ then my function becomes $\operatorname{e}^{-ix}+\operatorname{e}^{ix} \equiv 2\cos x$.
Why is $\alpha\operatorname{e}^{x}$ the only solution for all $n$?
Notice that $\lambda=1$ is a solution to $\lambda^n-1=0$ for all $n$ and so $\alpha\operatorname{e}^{x}$, where $\alpha$ is a constant, is a solution to $f^{(n)}=f$ for all $n$. Moreover, $\lambda=1$ is the only solution to all of $\lambda^n-1=0$ since if $\lambda$ is a solution to $\lambda^n-1=0$ for all $n$ then it must be a solution when $n=1$. However, when $n=1$, there is only a single solution: $\lambda = 1$.
A: $sin''''(x) = \sin x$
In fact, $f^{(n)}(x) = f(x)$ can be solved by $e^{kx}$ with $k^n = 1$, so $k = \cos \frac{ 2\pi k}{n} + i \sin \frac{ 2\pi k}{n} $.
