If I want to prove that for any scalar field $f:\;\mathbb{R}^3\to\mathbb{R}:$ $$\int_V \boldsymbol{\nabla} f\;\mathrm{d}V=\int_{\partial V} f\;\mathrm{d}\mathbf{S}$$ Can I apply the divergence theorem to $\mathbf{a}_1=(f,0,0),\;\mathbf{a}_2=(0,f,0),\;\mathbf{a}_3=(0,0,f)$ and then stack the equalities into a single vector? So using: $$\int_V \frac{\partial f}{\partial x_i}\mathrm{d}V=\int_{\partial V}f\cdot n_i\;\mathrm{d}S$$ ($n_i$ is the $i$th component in the outward normal $\mathbf{n}$) can I deduce:
$$\int_V \left(\frac{\partial f}{\partial x_1},\frac{\partial f}{\partial x_2},\frac{\partial f}{\partial x_3}\right)\mathrm{d}V=\int_{\partial V}f\Big(n_1,n_2,n_3\Big)\mathrm{d}S$$
$$\Rightarrow \int_V \boldsymbol{\nabla} f\;\mathrm{d}V=\int_{\partial V} f\;\mathrm{d}\mathbf{S}?$$
Let $\mathbf c$ be some constant vector. Taking the divergence of $\mathbf c f$,
$$\nabla\cdot(\mathbf c f)=\mathbf c \cdot \nabla f+f\nabla\cdot\mathbf c\\ =\mathbf c \cdot \nabla f,$$
since the divergence of a constant vector is zero.
Applying the divergence theorem to $\mathbf c f$,
$$\begin{align} \int_V \mathbf c \cdot \nabla f \,\mathrm{d} V&=\int_{\partial V}\mathbf c f\cdot\mathrm{d} \mathbf{S}\\ \implies \mathbf c \cdot \int_V \nabla f \,\mathrm{d} V&=\mathbf c \cdot \int_{\partial V}f\,\mathrm{d} \mathbf{S}\\ \implies 0&=\mathbf{c}\cdot\left(\int_V \nabla f \,\mathrm{d} V - \int_{\partial V}f\,\mathrm{d} \mathbf{S}\right). \end{align}$$
Since the previous equation holds for any constant vector $\mathbf{c}$, it follows that:
$$\int_V \nabla f \,\mathrm{d} V = \int_{\partial V}f\,\mathrm{d} \mathbf{S}.$$