If $f: \mathbb{R^n} \rightarrow \mathbb{R^m}$ be a function with $\nabla f(x) = 0$ for all $x$, then $f$ is constant. Suppose that $f$ is differentiable and that $\nabla f(x) = 0 \forall x \in \mathbb{R}^n$. Prove that there is a $v \in \mathbb{R}$ such that $f(x) = v \\ \forall x \in \mathbb{R^n}$
could anyone offer some guidance, I am completely stuck.
here's what I have thus far:
Fix $x,y \in \mathbb{R^n}$. Let $p:[0,1] \rightarrow \mathbb{R^n}$, defined by: 
$p(t) = x+t(y-x)$, so that $f_{i} \cdot p:[0,1] \rightarrow \mathbb{R}$.
We know that $\bigtriangledown f(x) = \textrm{ [}\bigtriangledown f_{1}(x),\bigtriangledown f_{2}(x), ..... , \bigtriangledown f_{m}(x) \textrm{]} = 0$ .
So, the derivative of the composite function $(f_{i} \cdot p)' = (f' \cdot p) \cdot p' = 0$ ? Since we know that $\bigtriangledown f(x) = 0$
 A: Let $(x_1,\cdots, x_n),(y_1,\cdots, y_n) \in \mathbb{R}^n$.
Consider the "$n$" functions $f_{i,p}(h)=f(p_1,\cdots,p_i+h,\cdots, p_n)$. Each one of them is a function $f_{i,p}: \mathbb{R} \rightarrow \mathbb{R}$. It is clear that $f_{i,p}'(h)=D_if(p_1,\cdots ,p_i+h, \cdots, p_n),$ where $D_i$ is the partial derivative with respect to the $i-$th coordinate. $\nabla f=0$ shows that $f_{i,p}'=0$ for all $i,p$. Hence, each $f_{i,p}$ is constant. It is clear that $f_{i,(p_1,\cdots,p_i +j,\cdots, p_n)}(\cdot)=f_{i,(p_1,\cdots,p_i ,\cdots, p_n)}(\cdot +j)$.
Consider now $f_{1,(x_1,\cdots, x_n)}$, $f_{2,(y_1,x_2,\cdots,x_n)}$,  ... , $f_{n,(y_1,\cdots,y_{n-1}, x_n)}$.
A: It's enough to treat the case $m=1.$ (By the way, what does $\nabla f$ mean if $m>1$?)
The result holds when $n=1;$ this is just one variable stuff. Suppose it's true for $n.$ Then for every fixed $y \in \mathbb R, f(x_1,\dots, x_n,y)$ is constant as a function of $(x_1,\dots, x_n)\in \mathbb R^n.$  Thus $f(x_1,\dots, x_n,y) = f(0,\dots,0,y)$ for all $(x_1,\dots, x_n,y) \in \mathbb R^{n+1}.$ But $y \to f(0,\dots,0,y) $ is also constant by the one variable result. It follows that $f$ is constant on $\mathbb R^{n+1}$ and we're done by induction.
