Find all functions $f$ satisfying a certain property How can we find all real-valued functions $f$ such that $f^{(n+1)}(x) = f^{(n)}(x)$? The question gives a hint which says: "If $n=2$, then we have $f(x) = ae^x+be^{-x}$, for some $a,b \in \mathbb{R}$. After proving this proceed by induction on $n$." I can see how to complete the proof assuming the hint, but how can I prove that fact?
Clarification: We are not allowed to assume that $f(x) = ae^x+be^{-x}$, but rather need to show this when $n = 2$. 

Take $f(x) = ae^x + be^{-x}$.  If $n$ is odd, then $f^{(n)}(x) = ae^x - be^{-x}$.  If $n$ is even, then $f^{(n)}(x) = ae^x + be^{-x}$, which shows that $b = 0$ (since the derivative of $e^{-x}$ depends on its order).  Therefore, $f(x) = ae^{x}$, where $a$ is any value.
I don't think there are any sophisticated method to prove that $f$ is of the form $ae^{x}$, where $a$ is an arbitrary constant.
