Stokes or homotopy? The problem states

Show that if $X$ is a simply connected manifold, then $\oint_{\gamma}\omega=0$ for all closed 1-forms  $\omega$ on X and all closed curves $\gamma$ in $X$. 

However I have answer it without the simply connected assumption, using Stokes Theorem, here is my proof:
We have $\gamma:S^1 \to X$, since $\omega$ is closed, then $d\omega=0$, so 
$$
\oint \omega = \int_{S^1} \gamma^*\omega \overset{Stokes}{=}\int_{B^2} d(\gamma^*\omega)=\int_{B^2} \gamma^*(d\omega)=\int_{B^2} 0=0
$$
I am wrong? I think I am because I am dropping an important assumption. 
I first try to answer it with the hint, which is to use that if $f_0$ and $f_1$ are homotopic maps then $\int_X f_0^* \omega = \int_X f_1^* \omega$ (I have already prove this Hint). This is what I got:
Since $X$ is a simply connected manifold I know that $\gamma$ is homotopic to a constant map $g:S^1 \to \left\{ x_0 \right\} \subset X$, then by the Hint
$$
\oint \omega = \int_{S^1} \gamma^*\omega =  \int_{S^1} g^*\omega
$$
BUT I CAN´T GO ANY FURTHER BECAUSE I AM HAVING TROUBLE ON COMPUTING $g^*\omega$.
 A: You are using that $\partial B^2 =  S^1$, but by doing so you assumed already that your $S^1$ is homotopically trivial (it bounds a disk). So you have used the condition (simply connectedness) already. Your first answer is correct (module the fact that you can find a smooth disk $B^2$)
For your second question, as $g$ is a constant map, then $g^*\omega = 0$. To see why, as your map is constant, there is no harm to assume that $M$ is an open set $U$ in $\mathbb R^n$. In general, if the map $g$ is given by 
$$g(t) = (f_1(t), \cdots f_n(t)), $$ 
and $\omega (x) = \alpha_1(x) dx^1 + \cdots \alpha_n(x) dx^n$, then 
$$g^* \omega (t) = \alpha_1 (g(t)) d f_1(t) + \cdots \alpha_n(g(t)) d f_n(t)$$
$$=\bigg(\alpha_1  (g(t)) f'_1(t) + \cdots \alpha_n(g(t)) f'_n(t) \bigg) dt$$
Now as $g$ is constant, $f'_i(t)=0$ for all $i$ and so $g^*\omega = 0$. 
A: It doesn't make sense to  integrate $d(\gamma^*\omega)$ on $B_2$ since $\gamma^*\omega$ is not defined on $B_2$ and consequently neither is $d(\gamma^*\omega)$.
Put differently an interior point of $B_2$ is not related in any way to any point of $X$.
Thus your use of Stokes is indeed unwarranted.
