All partial derivatives are 0. I know that for a function $f$ all of its partial derivatives are $0$.
Thus, $\frac{\partial f_i}{\partial x_j} = 0$ for any $i = 1, \dots, m$ and any $j = 1, \dots, n$. 
Is there any easy way to prove that $f$ is constant? The results seems obvious but I'm having a hard time expressing it in words explicitly why it's true.

It's not hard to see that given $\textbf{p,q}\in\mathbb{R}^M$:
$\hspace{2cm}\textbf{f(p)}-\textbf{f(q)}=\textbf{f}\bigl(f_1(\textbf{p})-f_1(\textbf{q}), f_2(\textbf{p})-f_2(\textbf{q}), ... f_N(\textbf{p})-f_N(\textbf{q})\bigr)$
$\hspace{4.5cm} =\bigl(\int_{\gamma} \nabla f_1(\mathbf{r})\cdot d\mathbf{r}, \int_{\gamma} \nabla f_2(\mathbf{r})\cdot d\mathbf{r}, ... \int_{\gamma} \nabla f_N(\mathbf{r})\cdot d\mathbf{r}\bigr)$
$\hspace{4.5cm}=0$
Since according to the Gradient Theorem, given any curve $\gamma$ with end points $\textbf{p},\textbf{q} \in \mathbb{R}^N$, we have:
$\hspace{6cm} f_i\left(\mathbf{p}\right)-f_i\left(\mathbf{q}\right) = \int_{\gamma} \nabla f_i(\mathbf{r})\cdot d\mathbf{r} $
and $\nabla f_i  = \frac{\partial f_i}{\partial x_1 }\mathbf{e}_1 + \cdots + \frac{\partial f_i}{\partial x_N }\mathbf{e}_N=0$ for all $i$, by assumption.
