Finding the solution to homogeneous linear equations with constant coefficients. This is what it says in my textbook, with no further detail as to why this is:
$$L(x) \equiv x^{(n)}+a_1x^{(n-1)}+\cdots+a_{n-1}x'+a_nx=0.$$
$$a_1 ,a_2,\ldots,a_n \in \mathbb R$$
According to Euler we search for the solution in the form of: 
$$x=e^{\lambda t}, \lambda \in \mathbb R$$
Does anyone know why this is exactly ?

An heuristic argument :
The $e^{\lambda t}$ are eigenvectors of the operator $L$, so it's natural to see if one of them is an eigenvector associated to the eigenvalue 0.
They are indeed eigenvectors because 
$$L(e^{\lambda t} ) = e^{\lambda t} \underbrace{ \left( \lambda^n + \sum_{k=1}^n a_k \lambda^{(n-k)} \right) }_{\gamma\in \mathbb{R}}$$
