If $f$ is differentiable everywhere and $|f(y)-f(x)| \leq |x-y|^n$ for all $n >1$, then $f^{\prime}(x)= 0 $ for all $x$. Suppose $f:\mathbb{R} \rightarrow \mathbb{R}$ is a differentiable function and satisfies $|f(x) - f(y)| \leq |x-y|^n$ for all $n > 1.$ Show that $f^{\prime}(x)=0$ for all $x\in \mathbb{R}.$
My first attempt: 
Choose $x < y$ such that $y - x < 1$. By the Mean Value Theorem on $[x,y]$, we have $\frac{f(y) - f(x)}{y-x} = f^{\prime}(c)$ for some $c \in (x,y).$ Note that we have $| f^{\prime}(c) | = \frac{|f(y) - f(x)|}{y-x} \leq |x-y|^{n-1}$. Since it holds for any $n > 1$, we have $\displaystyle\lim_{n \rightarrow \infty}|f^{\prime}(c)| \leq \displaystyle\lim_{n \rightarrow \infty}|x-y|^{n-1}.$ Since $y - x < 1,$ we have $\displaystyle\lim_{ \rightarrow \infty}|x-y|^{n-1} = 0$. Hence, $|f^{\prime}(c)| = \displaystyle\lim_{n \rightarrow \infty}|f^{\prime}(c)| = 0 .$ Therefore, $f^{\prime}(c) = 0$. Since it holds for any interval $[x,y]$ that shrinks to a point in it, we can have $f^{\prime}(x)=0$ for all $x \in \mathbb{R}.$
My second attempt: 
Since $f$ is differentiable everywhere, we have for any $x \in \mathbb{R}, f^{\prime}(x) = \displaystyle\lim_{y \rightarrow x }\frac{f(x)-f(y)}{x-y}.$ Therefore, $|f^{\prime}(x)| = \displaystyle\lim_{y \rightarrow x }\frac{|f(x)-f(y)|}{|x-y|} \leq \displaystyle\lim_{y \rightarrow x} |x-y|^{n-1} = 0.$ Hence, $|f^{\prime}(x)| = 0$, which implies that $f^{\prime}(x) = 0.$ Since it holds for any $x \in \mathbb{R}$, we have $f^{\prime}(x) = 0$ for any $x \in \mathbb{R}.$
Are my two attempts correct? I am quite doubtful about my first attempt on the choosing $x$ and $y$ part. 

Let's make it more general: if $f:{\mathbb R}\to {\mathbb R}$ is a function satisfying $|f(x) - f(y) |\le |x-y|^{1+\alpha}$ for some $\alpha>0$, then $f$ is constant. 
Let's see why.  Let $x=0$.  For any $n\in{\mathbb N}$, telescoping and triangle inequality give 
$$|f(y) - f(0)|\le \sum_{k=1}^n \left |f(\frac{y k}{n})-f(\frac{y (k-1)}{n})\right |=(*)$$ 
By assumption 
$$(*)  \le \sum_{k=1}^n |\frac{y}{n}|^{1+\alpha}=|y|^{1+\alpha} \frac{n}{n^{1+\alpha}}\to 0.$$ 
Therefore $f(y)=f(0)$ for all $y$. 
