We have a differential equation $$\frac{d\phi_\nu(t)}{dt}-\xi(\omega+\phi_{\nu-1}(t)-\phi_{\nu}(t))=0$$ When time is up, we have $\phi_\nu(0)=0$ for all values of $\nu \ge 1$, which is an integer environmental factor and also $\phi_0(t)=0$ at all times.
If we start from $\nu=1$ and recursively solve the equation for some values of $\nu$, we get $$\phi_\nu(t)=\omega\nu-\sum_{i=0}^{\nu-1}\frac{i+1}{(\nu-i-1)!}(\xi t)^{\nu-i-1}\omega e^{-\xi t}$$
But I only solved the equation for $\nu=1,\ldots,5$, and this formulation is true for those values. How can I prove this is the general form of $\phi$?
Constructing a solution from the first successive steps may be quite uncertain. Here is a more systematic method to deal with this equation.
Laplace transform allows you to "sweep the derivative under the carpet" in order to focus on the recurrence relation. Defining $$ \Phi_\nu(s) = \int_{0^-}^\infty \phi_\nu(t)e^{-st} \mathrm{d}t, $$ the original differential equation becomes $$ s\Phi_\nu(s) -\xi\left(\frac{\omega}{s} + \Phi_{\nu-1}(s) - \Phi_\nu(s)\right) = 0, $$ hence the following recurrence relation : $$ \Phi_\nu(s) = \frac{\Phi_{\nu-1}(s)}{s+\xi^{-1}} + \frac{\omega}{s(s+\xi^{-1})} $$ This nonhomogeneous first-order linear recurrence relation (with constant coefficients with respect to $\nu$) leads easily to $$ \Phi_\nu(s) = \frac{\Phi_0(s)}{(s+\xi^{-1})^\nu} + \frac{\omega}{s^2}\left(1-\frac{1}{(s+\xi^{-1})^\nu}\right) $$ Then you still need to take the inverse Laplace transform, but the result will depend on the initial term $\Phi_0(s)$ (that you didn't specify).