So let $$\dot{y}=y^n \quad \text{for} \ n \in \mathbb N$$
I need to find all solutions for every $n \in \mathbb N$ and I am not sure if I did it the right way.
$\dot{y}=y^n \Longleftrightarrow \int\frac{1}{y^n}dy = \int 1 dt $
So we can consider 2 cases
If $n=1 \implies y = e^{t+c} $ for a constant $c$
If $n \neq 1 \implies \frac{y^{-n+1}}{-n+1} = t+c \Longleftrightarrow y^{-n+1}=(-n+1)(t+c) \implies y = \sqrt[n-1]{\frac{1}{(t+c)(-n+1)}} $ if $ ((t+c)(-n+1)) \neq 0$
Am I missing something?
For $n=1$ it should be $y(t) = Ce^t$ for any constant $C$, because solving the integrals gives us $\ln|y|=t+c$, so $y=\pm e^ce^t$ and the case $c=0$ is obvious from $\dot{y}=y$. Right now your answer only covers $C>0$.
The case $n\neq 1$ (with $n$ odd) is only defined for $t<c$ due to even $k$-roots not being defined for negative input. For $n$ even you can just say that it is defined for $t\neq c$.