Find $f$, such that $\,f,f',\dots,f^{(n-1)}\,$ linearly independent and $\,f^{(n)}=f$ I am trying to find a function $f\in\mathcal{C}^\infty(\mathbb{R},\mathbb{C})$, satisfying the differential equation
$$
f^{(n)}=f,
$$
and with $\,f,f',\dots,f^{(n-1)}\,$ being linearly independent.
Could you give me some hints? 
Thanks.

The characteristic polynomial of the equation $x^{(n)}-x=0$ is $p(\zeta)=\zeta^n-1$,
with roots $1,\lambda,\ldots,\lambda^{n-1}$, where $\lambda=\exp(i\omega)$ with
$\omega=\dfrac{2\pi}{n}$. Hence, the functions
$$
f_k(t)=\exp \big(\lambda^k t\big), \,\,\,\text{where $k\in\{0,1,2,\ldots,n-1\}$}
$$
form a basis of the solution space of the equation $x^{(n)}-x=0$, i.e., the $f_k$'s are linearly independent and every solution of $x^{(n)}-x=0$ is a linear combination fo the $f_k$'s. Note also that $f_k^{(j)}(t)=\lambda^{kj}f_k(t)$.
Claim. If $a_0,\ldots,a_{n-1}$ are non-zero complex constants, 
and $f=\sum_{k=0}^{n-1}a_kf_k$, then the functions
$f,f',\ldots,f^{(n-1)}$ are linearly independent over $\mathbb C$.
Proof. We have that 
$$
f^{(j)}=\sum_{k=0}^{n-1}a_kf_k^{(j)}=\sum_{k=0}^{n-1}a_k\lambda^{kj}f_k.
$$
It $\sum_{j=0}^{n-1}c_jf^{(j)}=0$, for some $c_0,\ldots,c_{n-1}\in\mathbb C$, then
$$
0=\sum_{j=0}^{n-1}c_jf^{(j)}
=\sum_{j=0}^{n-1}c_j\left(\sum_{k=0}^{n-1}a_k\lambda^{kj}f_k\right)
=\sum_{k=0}^{n-1}a_k\left(\sum_{j=0}^{n-1}c_j\lambda^{kj}\right) f_k.
$$
Now as the $f_k$'s are linearly independent and the $a_k$'s non-zero, we obtain the system
$$
\sum_{j=0}^{n-1}\lambda^{kj}c_j=0, \quad k=0,\ldots,n-1,
$$
which is an $n\times n$ linear system with system matrix a Van der Monde matrix $$
A=\big(\lambda^{(j-1)(k-1)}\big)_{k,j=1}^n,
$$ 
which is invertible, since $\lambda^k\ne\lambda^j$, for $0\le k<j\le n-1$ - Note that
$\det A=\prod_{0\le k<j<n}(\lambda^{j}-\lambda^k)$.
Thus the above system has
a unique solution - the zero one - which means that the $f_j$'s are linearly independent. 
Remark. The inverse also holds, for if $a_k=0$, for some $k\in\{0,1,\ldots,n-1\}$, then
the coreesponding $f_j$'s would be linearly dependent.
