Motion equation of a free falling object, with air resistance

I solved it for fun, but a question raised.

$\displaystyle m \dot v =mg-\gamma v$. With the integrating factor method, I reached $\displaystyle v=\frac{gm}{\gamma}+e^{-\frac{\gamma t}{m}}C$.

If I set $\displaystyle v(0)=0$, I get $\displaystyle v=\frac{gm}{\gamma} \left [ 1- e^{-\frac{\gamma t}{m}} \right ]$.

So when $\displaystyle t\to \infty$, $\displaystyle v\to \frac{gm}{\gamma}$.

It surprises me a bit, it means that if one let fall 2 balls of different masses, the heavier one will fall faster than the less heavy one. I'm not sure this result is true, can you confirm it please?

Also it means that the terminal velocity is $\displaystyle \frac{gm}{\gamma}$, in other words, weight over the drag coefficient. This could make sense, the lesser the drag coefficient, the faster the object falls.