Proof that $\int_{\overline \Omega} \det f'(x) \, dm(x) = 0$

Let $\Omega \subset \mathbb R^n$ be an regular open subset and let $f = (f_1,\dots,f_n) \in C^\infty (U)$, where $U$ is an open subset of $\mathbb R^n$ that contains $\overline \Omega$. Let's suppose that $f_j = 0$ in $\partial \Omega$ for some $j = 1,...,n$. I want to proof that $$\int_{\overline \Omega} \det f'(x) \, dm(x) = 0$$

My idea was find a convenient $n-1$-form and apply Stoke's Theorem. Since, $$\det f'(x) = \sum_{k=1}^n (-1)^{j+k} \, \frac{\partial f_j}{\partial x_k} \, \frac{\partial(f_1,\dots, f_{j-1},f_{j+1},\dots,f_n)}{\partial(x_1,\dots,x_{k-1},x_{k+1},\dots,x_n)}$$

Let's denote $F_k = \frac{\partial(f_1,\dots, f_{j-1},f_{j+1},\dots,f_n)}{\partial(x_1,\dots,x_{k-1},x_{k+1},\dots,x_n)}$ for each $k = 1,...,n$. Defining $$\alpha = \sum_{k=1}^n (-1)^{j+1} f_j \, F_k \, dx_1\wedge\dots\wedge dx_{k-1} \wedge dx_{k+1}\wedge \dots \wedge dx_n$$ and, hence

$$d\alpha = \sum_{k=1}^n (-1)^{j+1} (-1)^{k-1} [\frac{\partial f_j}{\partial x_k} F_k + f_j \frac{\partial F_k}{\partial x_k}] dx_1\wedge\dots\wedge dx_n$$

$$\therefore \quad d\alpha = \det f'(x)\, dx_1\wedge\dots\wedge dx_n + \sum_{k=1}^n (-1)^{j+k} \, f_j \frac{\partial F_k}{\partial x_k} \, dx_1\wedge\dots\wedge dx_n$$

Using Stoke's Theorem,

$$\int_{\overline \Omega} d\alpha = \int_{\partial \Omega} \alpha$$ Since f_j = 0 in $\partial \Omega$, the right side must be zero. However, the left side is $$\int_{\overline \Omega} \det f'(x)\, dm(x) + \int_{\overline \Omega} \left [ \sum_{k=1}^n (-1)^{j+k} \, f_j \frac{\partial F_k}{\partial x_k} \right ] \, dm(x)$$

I need help to understand why the second term of the above expression must be zero. Thank you

HINT: Write $\det f'(x)\,dm(x) = df_1\wedge\dots\wedge df_n = \pm d(f_j\, df_1\wedge\dots\wedge\widehat{df_j}\wedge\dots\wedge df_n)$.