What I wish to prove is that for a Cauchy-Euler equation of order $n$, the substitution $x=e^{t}$ transforms it into a linear differential equation with constant coefficients. To put it as a theorem:
Consider the Euler equation of order $n$ \begin{equation} x^{n}y^{(n)}+a_{n-1}x^{n-1}y^{(n-1)}+\cdots+a_{1}xy'+a_{0}y=0 \end{equation} The substitution $x=e^{t}$ transforms this into a linear differential equations with constant coefficients.
What I have tried
I have been drawn to use a inductive proof. Having examined the first three derivatives I am lead to believe that the $n$:th derivative (with substitution $x=e^{t}$) can be written as $y^{(n)}=e^{-nt}\prod_{k=1}^{n}(D-k+1)y$, where $D$ denotes the differential operator.
For the base case $n=1$ we have \begin{equation} \frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}=\frac{dy}{dx}e^{t}=\frac{dy}{dx}x \iff\frac{1}{x}\frac{dy}{dt}=\frac{dy}{dx} \end{equation} which means that \begin{equation} a_{1}x\frac{dy}{dx}+a_{0}y=0 \iff a_{1}\frac{dy}{dt}+a_{0}y=0 \end{equation} As such, the base case clearly transforms into a linear differential equation.
Now, suppose that the substitution $x=e^{t}$ transforms the $n-1$ first derivatives into a linear combination. We wish to show that this implies that $n$:th derivative can also be transformed into a linear combination \begin{align} \frac{d^{n}y}{dx^{n}}&=\frac{dy}{dx}\left[ \frac{d^{n-1}y}{dx^{n-1}} \right] \\ &=\frac{d}{dx}\left[e^{-(n-1)t}\prod_{k=1}^{n-1}(D-k+1)y \right] \\ &=\frac{d/dt}{e^{t}}\left[e^{-(n-1)} \right] \\ &=e^{-t}\left[-(n-1)e^{-(n-1)t}\prod_{k=1}^{n-1}(D-k+1)y+e^{-(n-1)t}\frac{d}{dt}\prod_{k=1}^{n-1}(D-k+1)y \right] \\ &= \cdots \\ &=e^{-nt}\prod_{k=1}^{n}(D-k+1)y \end{align}
I am not completely sure as to how I should complete the dotted line; what exactly does the derivative of the product of differential operators become?