Magnetic Flux outside of a long solenoid

[lelouch_v1]

I have been reading Griffith's Introduction to Electodynamics and i am currently at the chapter about magnetostatics. There is an example about a long solenoid with n units per length and radius R that shows a way of finding the magnetic vector potential. The magnetic field inside the solenoid is
$$\textbf{B} = {\mu_0}nI{\hat{z}} \ \ , \ \ {\text{inside solenoid}}$$ and $$\textbf{B} = 0 \ \ , \ \ {\text{outside solenoid}}$$
What i cannot understand is that when dealing with the magnetic vector potential, Griffiths states that magnetic flux outside of the solenoid is
$$\int{\textbf{B}{\cdot}d{\textbf{a}}} = {\mu_0}nI({\pi}R^2)$$ and
$$\textbf{A}=\frac{\mu_0nIR^2}{2s}{\hat{\phi}} \ \ , \ \ s>=R$$
Why do we use the magnetic field that we have on the inside, when trying to find the magnetic flux on the outside?

 

[Vanadium 50]

Flux lines are neither created nor destroyed. If you have flux on the inside it needs to be returned on the outside.

 

[Meir Achuz]

B is continuous at the end of the solenoid, so B just inside the end equals B just outside.
However, B at the end of a solenoid is one half of its value at the middle, so the flux should be one half of what your equation gives. Your equation, B=0, is also wrong.